Comprehensive CBSE Question Bank in Mathematics-XII (Term-II)

Three Dimensional Geometry 259 Coordinates of D. Since ABCD is a parallelogram, so that diagonals AC and BD will bisect each other. ∴ Midpoint of AC = Midpoint of BD HGF JIK GFH KJI⇒ 4 + 1, 5 + 2 , 10 − 1 = 2 + x′ , 3 + y′ , 4 + z′ ...(1) 2 2 2 2 2 2 (1) ⇒ 2 + x′ = 5 ⇒ x′ = 5 – 2 = 3 2 2 (1) ⇒ 3 + y′ = 7 ⇒ y′ = 7 – 3 = 4 2 2 (1) ⇒ 4 + z′ = 9 ⇒ z′ = 9 – 4 = 5 2 2 ∴ Coordinates of D are (3, 4, 5). Example 13. Using vectors, show that the points A(–2, 3, 5), B(7, 0, –1), C(–3, –2, –5) and D(3, 4, 7) are such that AB and CD intersect at the point P(1, 2, 3). (A.I. CBSE 2012 C) Solution. Let →a = P.V. of A(–2, 3, 5) = – 2i + 3 j + 5k → = 7iˆ – kˆ b = P.V. of B(7, 0, –1) →c = P.V. of C(–3, –2, –5) = –3iˆ – 2 ˆj – 5kˆ and →d = P.V. of D(3, 4, 7) = 3iˆ + 4ˆj + 7kˆ Equation of AB is →r = →a + λ(→b – →a ) . C(–3, –2, –5) B(7, 0, –1) ⇒ →r = – 2iˆ + 3ˆj + 5kˆ + λ(7iˆ − kˆ + 2iˆ – 3ˆj – 5kˆ) P(1, 2, 3) ⇒ →r = – 2iˆ + 3 ˆj + 5kˆ + λ(9iˆ – 3 ˆj – 6kˆ) ...(1) A(–2, 3, 5) D(3, 4, 7) →→ →→ Equation of CD is r = c + μ( d – c ) . ⇒ →r = – 3iˆ – 2 ˆj – 5kˆ + μ(3iˆ + 4 ˆj + 7kˆ + 3iˆ + 2 ˆj + 5kˆ) ⇒ →r = – 3iˆ – 2 ˆj – 5kˆ + μ(6iˆ + 6 ˆj + 12kˆ) ...(2) Let (1) and (2) intersect at the point (x, y, z) with P.V. →r . ∴ –2iˆ + 3ˆj + 5kˆ + λ (9iˆ – 3ˆj – 6kˆ) = – 3iˆ – 2ˆj – 5kˆ + μ(6iˆ + 6 ˆj + 12kˆ) ⇒ (–2 + 9λ)iˆ + (3 – 3λ)ˆj + (5 – 6λ)kˆ = (–3 + 6μ)iˆ + (–2 + 6μ)ˆj + (–5 + 12μ)kˆ ⇒ –2 + 9λ = –3 + 6μ, 3 – 3λ = –2 + 6μ and 5 – 6λ = –5 + 12μ ⇒ 9λ – 6μ = –1, ...(3) 3λ + 6μ = 5, ...(4) 6λ + 12μ = 10 ...(5)

260 Mathematics—XII Solving (3) and (4), we get λ = 1/3, μ = 2/3. (5) ⇒ 6(1/3) + 12(2/3) = 10 ⇒ 2 + 8 = 10, which is true. ∴ Lines AB and CD are intersecting. Putting λ = 1/3 in (1), we get →r = – 2iˆ + 3 ˆj + 5kˆ + 1 (9iˆ – 3 ˆj – 6kˆ) = iˆ + 2 ˆj + 3kˆ . 3 ∴ The lines AB and CD intersect at the point P(1, 2, 3). Example 14. Show that the points whose position vectors given by 2iˆ + ˆj + 3kˆ, −4 iˆ + 3ˆj − kˆ and 5iˆ + 5kˆ are collinear. Solution. Let A, B and C be the points such that P.V. of A = 2iˆ + ˆj + 3kˆ , P.V. of B = −4iˆ + 3ˆj − kˆ and P.V. of C = 5iˆ + 5kˆ . ∴ The coordinates of A, B, C are (2, 1, 3), (–4, 3, –1), (5, 0, 5) respectively. d.r.’s of AB are –4 – 2, 3 – 1, –1 – 3 or –6, 2, –4 or 3, –1, 2. The line AB passes through A(2, 1, 3). ∴ The equations of AB are x − 2 = y−1 = z − 3 . 3 −1 2 Putting x = 5, y = 0, z = 5, we get 5 − 2 = 0−1 = 5 − 3 or 1 = 1 = 1, which is true. 3 −1 2 ∴ C lies on AB. ∴ The points with given position vectors are collinear. Example 15. Determine whether the following pair of lines intersect: → = iˆ + ˆj − kˆ + λ(3iˆ − ˆj ) and → = 4iˆ − kˆ + μ(2iˆ + 3kˆ) . r r (A.I. CBSE 2013 C; CBSE 2014) Solution. Given lines are →r = iˆ + ˆj − kˆ + λ(3iˆ − ˆj) ...(1) and →r = 4iˆ − kˆ + μ(2iˆ + 3kˆ) . ...(2) (1) ⇒ →r = →a + λ→b , where →a = iˆ + ˆj − kˆ and →b = 3iˆ − ˆj (2) ⇒ →r = →c + μ→d , where →c = 4iˆ − kˆ and →d = 2iˆ + 3kˆ ∴ →c − →a = (4iˆ − kˆ) − (iˆ + ˆj − kˆ) = 3iˆ − ˆj 3 −1 0 ∴ [→c − →a →b →d ] = (→c − →a ) ⋅ (→b × →d ) = 3 −1 0 = 3(–3) + 1(9) = 0 2 03 ∴ The lines are coplanar. The vectors →b = 3iˆ − ˆj and →d = 2iˆ + 3kˆ are not parallel, so that the lines are not parallel. ∴ The given lines must be intersecting.

Three Dimensional Geometry 261 Example 16. Show that the lines →r = 3iˆ + 2ˆj – 4kˆ + λ(iˆ + 2ˆj + 2kˆ) and →r = 5iˆ – 2ˆj + μ(3iˆ + 2ˆj + 6kˆ) are intersecting. Hence find their point of intersection. (A.I. CBSE 2013) Solution. Given lines are → = 3iˆ + 2 ˆj – 4kˆ + λ(iˆ + 2 ˆj + 2kˆ) ...(1) r and →r = 5iˆ – 2 ˆj + μ(3iˆ + 2 ˆj + 6kˆ) ...(2) (1) ⇒ →r = (3 + λ)iˆ + (2 + 2λ)ˆj + (– 4 + 2λ)kˆ (2) ⇒ →r = (5 + 3μ)iˆ + (–2 + 2μ)ˆj + 6μkˆ If the given lines are intersecting, they must have a common point. ∴ For some values of λ and μ, we have (3 + λ)iˆ + (2 + 2λ)ˆj + (–4 + 2λ)kˆ = (5 + 3μ)iˆ + (–2 + 2μ)ˆj + 6μkˆ ⇒ 3 + λ = 5 + 3μ, 2 + 2λ = –2 + 2μ and –4 + 2λ = 6μ ∴ λ – 3μ = 2, ...(3) 2λ – 2μ = –4, ...(4) 2λ – 6μ = 4 ...(5) Solving (4) and (5), we have λ = –4 and μ = –2. (3) ⇒ –4 – 3(–2) = 2 ∴ Values of λ and μ satisfy (3). ∴ The given lines are intersecting. Putting λ = – 4 in (1), we get →r = 3iˆ + 2 ˆj – 4kˆ + (–4)(iˆ + 2 ˆj + 2kˆ) = – iˆ – 6ˆj – 12kˆ ∴ P.V. of point of intersection is – iˆ – 6ˆj – 12kˆ . ∴ The point of intersection is (–1, –6, –12). Remark. The value of μ = –2 in (2) would give the same point of intersection. Example 17. Show that the lines x + 1 = y + 3 = z + 5 and 3 5 7 x − 2 = y−4 = z–6 intersect. Also, find their point of intersection. 1 3 5 (CBSE 2014) Solution. Given lines are x + 1 = y + 3 = z+5 = λ, say ...(1) 3 5 7 and x − 2 = y − 4 = z–6 = μ, say ...(2) 1 3 5 (1) ⇒ x = 3λ – 1, y = 5λ – 3, z = 7λ – 5 ∴ Any point on (1) is (3λ – 1, 5λ – 3, 7λ – 5). (2) ⇒ x = μ + 2, y = 3μ + 4, z = 5μ + 6 ∴ Any point on (2) is (μ + 2, 3μ + 4, 5μ + 6). If the given lines intersect, they must have a common point.

262 Mathematics—XII ∴ For some values of λ and μ, we have 3λ – 1 = μ + 2, 5λ – 3 = 3μ + 4 and 7λ – 5 = 5μ + 6 ∴ 3λ – μ = 3, ...(3) 5λ – 3μ = 7, ...(4) 7λ – 5μ = 11 ...(5) Solving (3) and (4), we get λ = 1/2, μ = –3/2. (5) ⇒ 7(1/2) – 5(–3/2) = 11 ∴ Values of λ and μ satisfy (5). ∴ The given lines are intersecting. Putting λ = 1/2 in (3λ – 1, 5λ – 3, 7λ – 5), the point of intersection is (1/2, –1/2, –3/2). Example 18. The vector equations of two lines are: →r = (1 − t)iˆ + (t − 2)ˆj + (3 − 2t)kˆ and →r = (s + 1)iˆ + (2s − 1)ˆj – (2s + 1)kˆ . Find the shortest distance between these lines. (NCERT) Solution. The given lines are → = (1 − t)iˆ + (t − 2)ˆj + (3 − 2t)kˆ ...(1) r and → = (s + 1)iˆ + (2s − 1)ˆj – (2s + 1)kˆ . ...(2) r (1) ⇒ → = iˆ − 2ˆj + 3kˆ + t( − iˆ + ˆj − 2kˆ) r ∴ →r = →a + t →b , where →a = iˆ − 2ˆj + 3kˆ and →b = − iˆ + ˆj − 2kˆ (2) ⇒ →r = iˆ − ˆj − kˆ + s(iˆ + 2ˆj − 2kˆ) ∴ → = → + s → , where → = iˆ − ˆj − kˆ r c d c and → = iˆ + 2 ˆj − 2kˆ d SD between lines = |(→c − →a ) ⋅ (→b × →d )| |→b × →d | Now, iˆ ˆj kˆ →b × →d = −1 1 −2 = iˆ(−2 + 4) − ˆj(2 + 2) + kˆ(−2 − 1) = 2iˆ − 4ˆj − 3kˆ 1 2 −2 ∴ |→b × →d|= 4 + 16 + 9 = 29 Also, →c − →a = (iˆ − ˆj − kˆ) − (iˆ − 2 ˆj + 3kˆ) = ˆj − 4kˆ ∴ → − → ⋅ → × → = ( ˆj − 4kˆ) . (2iˆ − 4 ˆj − 3kˆ) (c a) (b d) = (0)(2) + (1)(–4) + (–4)(–3) = 0 – 4 + 12 = 8 → → →→ |8| |( c − a)⋅(b × d )| = 29 8 units. ∴ SD = = 29 →→ |b × d|

Three Dimensional Geometry 263 Example 19. Find the SD between the lines: x − 3 = y−8 = z−3 and x+3 y+7 = z − 6 . 3 −1 1 −3 = 2 4 Also, find the equations of the line of shortest distance. Solution The given lines are x − 3 = y−8 = z − 3 ...(1) and x+3 = y + 7 = z − 6 . ...(2) 3 −1 1 −3 2 4 The line (1) passes through A(3, 8, 3) 2 and has d.r.’s 3, –1, 1. The line (2) passes B(–3, –7, 6) through B(–3, –7, 6) and has d.r.’s Q –3, 2, 4. Let PQ be the line of shortest distance –3, 2, 4 between the lines with P and Q on the lines (1) and (2) respectively. Let P = (3λ1 + 3, – λ1 + 8, λ1 + 3) 3, –1, 1 1 and Q = (–3λ2 – 3, 2λ2 – 7, 4λ2 + 6) P A(3, 8, 3) for some scalars λ1 and λ2. d.r.’s of PQ are (–3λ2 – 3) – (3λ1 + 3), (2λ2 – 7) – (– λ1 + 8), (4λ2 + 6) – (λ1 + 3) or –3λ2 – 3λ1 – 6, 2λ2 + λ1 – 15, 4λ2 – λ1 + 3. PQ is ⊥ on line (1). ∴ 3(– 3λ2 – 3λ1 – 6) – 1(2λ2 + λ1 – 15) + 1(4λ2 – λ1 + 3) = 0 ⇒ –7λ2 – 11λ1 = 0 ...(3) PQ is ⊥ on line (2). ∴ – 3(– 3λ2 – 3λ1 – 6) + 2(2λ2 + λ1 – 15) + 4(4λ2 – λ1 + 3) = 0 ⇒ 29λ2 + 7λ1 = 0 ...(4) (3) and (4) imply λ1 = 0 and λ2 = 0. ∴ P = (0 + 3, 0 + 8, 0 + 3) = (3, 8, 3) and Q = (0 – 3, 0 – 7, 0 + 6) = (–3, –7, 6) ∴ Shortest Distance, PQ = (3 + 3)2 + (8 + 7)2 + (6 − 3)3 = 3 30 units. d.r.’s of PQ are –3 – 3, –7 – 8, 6 – 3 or –6, –15, 3 or 2, 5, –1. ∴ The equations of PQ i.e. of the line of shortest distance are x − 3 = y − 8 = z−3 . [∵ (3, 8, 3) is on PQ] 2 5 −1 Example 20. Find the shortest distance and the equation of the line of shortest distance between the lines: →r = 3i + 5j + 7k + λ(i − 2j + k) and →r = − iˆ − ˆj − kˆ + μ(7iˆ − 6ˆj + kˆ) . The given lines are Solution. →r = 3iˆ + 5 ˆj + 7kˆ + λ(iˆ − 2 ˆj + kˆ) ...(1) and →r = − iˆ − ˆj − kˆ + μ(7iˆ − 6ˆj + kˆ) . ...(2) Line (1) passes through the point (3, 5, 7) with position 3iˆ + 5ˆj + 7kˆ and has d.r.’s 1, –2, 1, because it is parallel to the vector iˆ − 2ˆj + kˆ .

264 Mathematics—XII ∴ The equation (1) in cartesian form is 2 x − 3 = y−5 = z − 7 ...(3) 1 −2 1 Similarly, the equation (2) in cartesian form (–1, –1, –1) Q is 7, –6, 1 x + 1 = y+1 = z + 1 ...(4) 7 −6 1 Let PQ be the line of shortest distance 1, –2, 1 1 between the lines with P and Q on the P (3, 5, 7) lines (1) and (2), respectively. Let P = (λ1 + 3, – 2λ1 + 5, λ1 + 7) and Q = (7μ1 – 1, – 6μ1 – 1, μ1 – 1) for some scalars λ1 and μ1. d.r.’s of PQ are (7μ1 – 1) – (λ1 + 3), (– 6μ1 – 1) – (– 2λ1 + 5), (μ1 – 1) – (λ1 + 7) or 7μ1 – λ1 – 4, – 6μ1 + 2λ1 – 6, μ1 – λ1 – 8 PQ is ⊥ on line (1) ∴ (7μ1 – λ1 – 4) – 2(–6μ1 + 2λ1 – 6) + 1(μ1 – λ1 – 8) = 0 ⇒ 10μ1 – 3λ1 = 0 ...(5) PQ is ⊥ on line (2) ∴ 7(7μ1 – λ1 – 4) – 6(–6μ1 + 2λ1 – 6) + 1(μ1 – λ1 – 8) = 0 ⇒ 43μ1 – 10λ1 = 0 ...(6) Solving (5) and (6), we get λ1 = 0 and μ1 = 0. ∴ P = (0 + 3, 0 + 5, 0 + 7) = (3, 5, 7) and Q = (0 – 1, 0 – 1, 0 – 1) = (–1, –1, –1) ∴ Shortest distance, PQ = (−1 − 3)2 + (−1 − 5)2 + (−1 − 7)2 = 16 + 36 + 64 = 116 = 2 29 units. Now P→Q = P.V. of Q – P.V. of P = (− iˆ − ˆj − kˆ) – (3iˆ + 5ˆj + 7kˆ) = − 4iˆ – 6ˆj − 8kˆ ∴ The equation of the line PQ is → = P.V. of P + λ P→Q r or → = 3iˆ + 5 jˆ + 7 kˆ + λ(− 4iˆ − 6 jˆ − 8kˆ ) . r PRACTICE EXERCISE 5.1 Very Short/Short Answer Type Questions (NCERT) 1. Find the d.c.’s of the X-axis, Y-axis and Z axis. 2. Find the direction cosines of a line which makes the following angles with coordinate axes: (i) 90º, 60º, 30º (NCERT) (ii) 90°, 135°, 45°. (NCERT) 3. Find the d.c.’s of the line whose d.r.’s are: (i) –18, 12, –4 (NCERT) (ii) 2, –1, –2. (NCERT; CBSE 2012)

Three Dimensional Geometry 265 4. (i) A line makes angles 90° with X-axis, 60° with Y-axis. Find the angle made by it with Z-axis. (CBSE 2015, 2017) (ii) If a line makes angles π/4 with each of X-axis and Y-axis, then what angle does it make with the Z-axis ? 5. The cartesian equations of a line are 3 – x = y + 4 = 2z – 6 . Find the vector equation of 5 7 4 the line. (A.I. CBSE 2014) 6. (i) Find the vector and cartesian equations of the straight line passing through the point (5, 2, –4) and parallel to the vector 3iˆ + 2ˆj − 8kˆ . (NCERT) (ii) Find the cartesian equations of a line parallel to the z-axis and passing through the point (α, β, γ). (A.I. CBSE 2014 C) 7. (i) Find the cartesian equations of the line which passes through the point (–2, 4, –5) and parallel to the line given by x +3 = 4– y = z+8 . (NCERT; CBSE 2013) 3 5 6 (ii) Find the equation of the line which passes through the point (1, 2, 3) and parallel to the vector 3iˆ + 2ˆj − 2kˆ . (NCERT) 8. A line passes through the point with position vectors 2iˆ − 3ˆj + 4kˆ and makes angles 60°, 120° and 45° with x, y and z-axis respectively. Find the equation of the line in the cartesian form. (CBSE 2016 C) 9. (i) Find the direction cosines of the line 4 – x = y = 1 – z . (CBSE 2013 C) 2 6 3 (ii) The equations of a line are 5x – 3 = 15y + 7 = 3 – 10z. Write the direction cosines of the line. (A.I. CBSE 2015) 10. (i) Find the direction cosines of the line x + 2 = 2y − 7 = 5 − z . Also, find the vector 2 6 6 equation of the line through the point A(–1, 2, 3) and parallel to the given line. (CBSE 2014 C) (ii) Find the vector equation of the line passing through the point (1, 2, 3) and parallel to the line x−1 = 1− y = 3−z . (A.I. CBSE 2016) −2 3 −4 11. (i) Find the cartesian and vector equations of the line which passes through the point (–2, 4, –5) and parallel to the line given by x+3 = y−4 = 8−z . (CBSE 2018 SP) 3 5 −6 (ii) Find the vector equation of the line passing through the point A(1, 2, –1) and parallel to the line 5x – 25 = 14 – 7y = 35z. (CBSE 2017) 12. (i) Find the vector equation of the line through the points (–1, 0, 2) and (3, 4, 6). (ii) Find the vector equation of the line through the points P(1, 6, 9) and Q(0, –3, 2). (NCERT) 13. (i) Find the vector and cartesian equations of the line that passes through the origin and (5, –2, 3). (NCERT) (ii) Find the vector and cartesian equations of the line that passes through the points (3, –2, –5) and (3, –2, 6). (NCERT) 14. Find the coordinates of the point where the line through A(3, 4, 1) and B(5, 1, 6) crosses the (i) xy-plane (ii) yz-plane (iii) xz-plane. (NCERT ; A.I. CBSE 2012) 265

266 Mathematics—XII Long Answer-I Type Questions 1. Find the d.c.’s of the line passing through the points: (i) (1, 0, 0) and (0, 1, 1) (ii) (–3, 6, 7) and (1, 4, 7) (iii) (2, 3, 5) and (–1, 2, 4) [NCERT (EP)] (iv) (–2, 4, –5) and (1, 2, 3). (NCERT) 2. If a line makes angles α, β, γ with the positive directions of axes, then show that: (i) sin2 α + sin2 β + sin2 γ = 2 (CBSE 2015 C) (ii) cos 2α + cos 2β + cos 2γ + 1 = 0. 3. Find the direction cosines of the sides of the triangle whose vertices are (3, 5, – 4), (–1, 1, 2) and (–5, –5, –2). (NCERT) 4. Show that the points with position vectors −iˆ + 4ˆj − 2kˆ, 2iˆ − 2ˆj + kˆ and 2ˆj − kˆ are collinear. 5. Show that the points with position vectors −2iˆ + 3ˆj + 5kˆ, iˆ + 2ˆj + 3kˆ, 7iˆ − kˆ are collinear. 6. If the points (–1, 3, 2), (–4, 2, –2) and (5, 5, λ) are collinear, then find the value of λ. 7. Show that the lines x+3 = y−1 = z−5 and x+1 = y−2 = z−5 are coplanar. −3 1 5 −1 2 5 (NCERT) 8. Show that the lines: x−a+d = y−a = z−a−d and x−b+c = y−b = z −b−c are coplanar. (NCERT) α−δ α α+δ β−γ β β+γ 9. Show that the lines →r = iˆ + ˆj − kˆ + λ(3iˆ − ˆj) and →r = 4iˆ − kˆ + μ(2iˆ + kˆ) intersect. Also, find their point of intersection. (CBSE 2014) 10. Show that the lines →r = 3iˆ + 2ˆj − 4kˆ + λ(iˆ + 2ˆj + 2 kˆ) and →r = 5iˆ − 2ˆj + μ(3iˆ + 2ˆj + 6kˆ) are intersecting. Hence, find their point of intersection. (A.I. CBSE 2013) 11. Show that the lines x − 1 = y − 2 = z − 3 and x − 4 = y − 1 = z intersect. Find their 2 3 4 5 2 point of intersection. [NCERT (EP)] 12. Show that the line through A(0, –1, –1) and B(4, 5, 1) intersect the line through C(3, 9, 4) and D(–4, 4, 4). Also, find their point of intersection. [NCERT (EP)] Long Answer-II Type Question 1. Find the shortest distance between the lines: → = iˆ + ˆj + λ(2iˆ − ˆj + kˆ) and → = 2iˆ + ˆj − kˆ + μ(3iˆ − 5 ˆj + 2kˆ) . (NCERT; CBSE 2014) r r 2. The vector equations of two lines are: → = iˆ + 2 ˆj + 3kˆ + λ (iˆ − 3 ˆj + 2kˆ) and → 4iˆ + 5 ˆj + 6kˆ + μ(2iˆ + 3 ˆj + kˆ) . r r= Find the shortest distance between the above lines. (NCERT; CBSE 2014 C) 3. Find the shortest distance between the lines: →r = iˆ + 2 ˆj + kˆ + λ(iˆ − ˆj + kˆ) and →r = 2iˆ − ˆj − kˆ + μ(2iˆ + ˆj + 2kˆ) . (NCERT; CBSE 2015C)

Three Dimensional Geometry 267 4. Find the shortest distance between the lines: → = 6iˆ + 2 ˆj + 2kˆ + λ(iˆ − 2 ˆj + 2kˆ) and → = − 4iˆ − kˆ + μ(3iˆ − 2 ˆj − 2kˆ) .(NCERT; CBSE 2013) r r 5. Find the shortest distance between the lines: →r = 2iˆ − 5 ˆj + kˆ + λ(3iˆ + 2 ˆj + 6kˆ) and →r = 7iˆ − 6kˆ + μ(iˆ + 2 ˆj + 2kˆ) . (A.I. CBSE 2015 C) 6. Define skew-lines. Using only vector approach, find the shortest distance between the following two skew-lines: →r = (8 + 3λ)iˆ − (9 + 16λ)ˆj + (10 + 7λ)kˆ and →r = 15iˆ + 29ˆj + 5kˆ + λ(3iˆ + 8ˆj − 5kˆ) . [NCERT (EP) ; CBSE 2018 SP] 7. By computing the shortest distance between the following pair of lines, determine whether they intersect or not ? →r = iˆ – ˆj + λ (2iˆ – kˆ) and →r = 2iˆ – ˆj + μ (iˆ – ˆj – kˆ) . (CBSE 2012 C) 8. Find the shortest distance between the lines: → = (t + 1) iˆ + (2 – t) ˆj + (1 – t) kˆ and → = (2s + 2) iˆ – (1 – s) ˆj + (2s – 1) kˆ . r r (CBSE 2016 C) 9. Find the shortest distance between the following pair of skew lines: x − 1 = 2 − y = z + 1 and x+2 = y − 3 = z . (CBSE 2016 SP) 2 3 4 −1 2 3 10. Find the shortest distance between the lines: x −3 = y−5 = z−7 and x+1 = y+1 = z + 1 . (NCERT ; CBSE 2014) 1 −2 1 7 −6 1 11. Find the shortest distance between the lines: x – 1 = y – 2 = z – 3 and x – 2 = y – 3 = z – 5 . (A.I. CBSE 2012 C; CBSE 2018 C) 2 3 4 3 4 5 12. Find the shortest distance between the lines through the points A(6, 2, 2) and A′(–4, 0, –1) in the directions 1, –2, 2 and 3, –2, –2 respectively. Answers Very Short/Short Answer Type Questions 1. 1, 0, 0 ; 0, 1, 0 ; 0, 0, 1 2. (i) 0, 1 , 3 (ii) 0, – 1 , 1 2 2 2 2 3. (i) 9 , 6 , 2 (ii) 2 ,– 1 , – 2 4. (i) 30° or 150° (ii) π/2 11 11 11 3 3 3 5. → = 3iˆ – 4 ˆj + 3kˆ + λ(– 5iˆ + 7 ˆj + 2kˆ) r 6. (i) →r = 5iˆ + 2 ˆj − 4kˆ + λ(3iˆ + 2 ˆj − 8kˆ), x − 5 = y −2 = z+4 ; (ii) x − α = y − β = z − γ 3 2 −8 0 0 1 7. (i) x + 2 = y−4 = z + 5 (ii) →r = iˆ + 2 ˆj + 3kˆ + λ(3iˆ + 2 ˆj − 2kˆ) 3 –5 6

268 Mathematics—XII 8. x − 2 = y+3 = z −4 9. (i) 2 , – 6 , 3 (ii) 6 , 2 , − 3 1 −1 2 7 7 7 7 7 7 10. (i) 2 , 3 , − 6 ; →r = − iˆ + 2 ˆj + 3kˆ + λ(2iˆ + 3 ˆj − 6kˆ) (ii) → = iˆ + 2 ˆj + 3kˆ + λ(2iˆ + 3 ˆj − 4kˆ) 7 7 7 r 11. (i) x+2 = y−4 = z + 5 , →r = – 2iˆ + 4ˆj − 5kˆ + λ (3iˆ + 5ˆj + 6kˆ) 3 5 6 → = iˆ + 2ˆj − kˆ + λ(7iˆ − 5ˆj + kˆ) (ii) r 12. (i) → = – iˆ + 2kˆ + λ(iˆ + ˆj + kˆ) (ii) → = iˆ + 6 ˆj + 9kˆ + λ(−iˆ − 9 ˆj − 7 kˆ ) r r 13. (i) → = λ(5iˆ − 2 ˆj + 3kˆ) , x = y = z (ii) → = 3iˆ − 2 ˆj − 5kˆ + λkˆ , x − 3 = y + 2 = z + 5 5 −2 3 0 0 1 r r 14. (i) ⎛ 13 , 23 , 0 ⎞ (ii) ⎛ 0, 17 , − 13 ⎞ (iii) ⎛ 17 , 0, 23 ⎞ ⎜⎝ 5 5 ⎟⎠ ⎜⎝ 2 2 ⎠⎟ ⎜⎝ 3 3 ⎠⎟ Long Answer-I Type Questions 1. (i) – 1 , 1 , 1 (ii) 2 , − 1 , 0 3 3 3 5 5 (iii) 3 , 1 , 1 (iv) 3 , − 2 , 8 11 11 11 77 77 77 3. 2, 2 ,– 3 ; 2 , 3 , 2 ; 4 , 5 , − 1 17 17 17 17 17 17 42 42 42 6. λ = 10. 9. (4, 0, –1) 10. (–1, –6, –12) 11. (–1, –1, –1) 12. (10, 14, 4) Long Answer-II Type Question 1. 10 59/59 units 2. 3 19/19 units 3. 3 2 /2 units 4. 9 units 5. 17 5/5 units 6. 14 units 7. 14/14 units, No 8. 5 26/26 units 9. 7 390/65 units 10. 2 29 units 11. 6/6 units 12. 9 units. HINTS/SOLUTIONS Long Answer-I Type Questions (b − c) − (a − d) b − a (b + c) − (a + d) 8. Use: α − δ α α + δ β−γ β β+γ b − c − a + d b − a 2b − 2a = α−δ α 2α = 0. (Operating C3 → C3 + C1) β−γ β 2β

Three Dimensional Geometry 269 B. PLANE IN SPACE Example 1. Find the vector equation of the plane which is at a distance of 5 units from the Solution. origin and perpendicular to the line, directed toward the plane, with d.c.’s 1 , 2 , – 3 . 14 14 14 Here l = 1 , m = 2 , n = – 3 , p = 5. We know that the equation of 14 14 14 the plane which is perpendicular to the line, directed toward the plane, with d.c.’s l, m, n and at a distance p from the origin is lx + my + nz = p. ∴ The equation of the required plane is 1 x+ 2 y + ⎛ − 3 ⎞ z = 5. 14 14 ⎝⎜ 14 ⎠⎟ ⇒ x + 2y – 3z = 5 14 ⇒ (iˆ + 2ˆj − 3kˆ) ⋅ (xiˆ + yˆj + zkˆ) = 5 14 ⇒ (iˆ + 2 jˆ − 3kˆ ) ⋅ → = 5 14 . r This is the vector equation of the required plane. Example 2. Find the equation of the plane which is at a distance 3 3 units from the origin Solution. and the normal to which is making equal acute angles with the positive coordinate Example 3. axes. [NECRT (EP)] Let the normal directed toward the required plane make angle α with each axis. Thus, α is an acute angle. ∴ The d.c.’s are cos α, cos α, cos α. ⇒ cos2 α + cos2 α + cos2 α = 1 ⇒ cos α = ± 1 3 ⇒ cos α = 1 (∵ α is acute) 3 ∴ The d.c.’s. of the normal directed toward the plane are 1 , 1 , 1 . 333 Also, p= 3 3. ∴ Using lx + my + nz = p, the equation of the required plane is 1 x+ 1 y+ 1 z= 3 3 333 i.e., x + y + z = 9. Find the direction cosines of the perpendicular from the origin towards the plane (iˆ + 2ˆj − 2kˆ) ⋅→r + 18 = 0. Solution. We have (iˆ + 2 ˆj − 2kˆ ) ⋅ → + 18 = 0. ⇒ r ⇒ (iˆ + 2ˆj − 2kˆ) ⋅ (xiˆ + yˆj + zkˆ) + 18 = 0 x + 2y – 2z + 18 = 0 ⇒ x + 2y – 2z = –18

270 Mathematics—XII ⇒ – x – 2y + 2z = 18 ...(1) (To make RHS +ve) Now, (−1)2 + (−2)2 + (2)2 = 1 + 4 + 4 = 3 Dividing (1) by 3, we get – 1 x − 2 y + 2 z = 6. 3 3 3 This is the normal form of the given equation. ∴ The d.c.’s of the perpendicular from the origin towards the plane are – 1 , − 2 , 2 . 3 3 3 Example 4. Find the distance of the plane 2x – 3y + 4z – 6 = 0 from the origin. (NCERT) Solution. We have 2x – 3y + 4z – 6 = 0. ⇒ 2x – 3y + 4z = 6 ...(1) (Note that RHS is already +ve) Now, (2)2 + (–3)2 + (4)2 = 4 + 9 + 16 = 29 Dividing (1) by 29 , the normal form of (1) is 2 x− 3 y + 4 z = 6 . Here p = 6 . 29 29 29 29 29 ∴ The distance of the given plane from the origin is 6 units. 29 Example 5. (i) Find the cartesian form of the plane → ⋅ (3iˆ − ˆj + 7kˆ) = 8. Solution. r Example 6. Solution. (ii) Find the vectorial form of the plane 2x + 9y – 3z + 21 = 0. (i) Given plane is →r ⋅ (3iˆ − ˆj + 7kˆ) = 8. ...(1) Let (x, y, z) be a general point on (1). ∴ →r = xiˆ + yˆj + zkˆ ∴ (1) ⇒ (xiˆ + yˆj + zkˆ) ⋅ (3iˆ − ˆj + 7kˆ) = 8 ⇒ 3x – y + 7z = 8. ...(1) This is the required cartesian form. (ii) Given plane is 2x + 9y – 3z + 21 = 0. Let → be the position vector of a general point (x, y, z ) on (1). r ∴ → = xiˆ + yˆj + zkˆ r (1) ⇒ (2iˆ + 9ˆj − 3kˆ) ⋅ (xiˆ + yˆj + zkˆ) = –21 ⇒ (2iˆ + 9jˆ − 3kˆ) ⋅ →r = –21. This is the required vectorial form. A variable plane passes through a fixed point (1, 2, 3). Show that the locus of the foot of the perpendicular drawn from the origin to this plane is given by the equationx2 + y2 + z2 – x – 2y – 3z = 0. Let P(α, β, γ) be the foot of the perpendicular drawn from the origin to the variable plane.

Three Dimensional Geometry 271 ∴ d.r.’s of OP are α – 0, β – 0, γ – 0 O(Origin) or α, β, γ and d.r.’s of AP are α – 1, β – 2, γ – 3. Since OP ⊥ AP, we have α(α – 1) + β(β – 2) + γ(γ – 3) = 0. P (a, b, g) ⇒ α2 + β2 + γ2 – α – 2β – 3γ = 0 ∴ The locus of the point P(α, β, γ) is A x2 + y2 + z2 – x – 2y – 3z = 0. (1, 2, 3) Example 7. Find the equation of the plane that Solution. bisects the line joining the points (1, 2, 3) and (3, 4, 5) and is at right A P (x, y, z) Example 8. angle to the line. (1, 2, 3) Solution. Let A(1, 2, 3) and B(3, 4, 5) be the MB given points. Let M be the midpoint (3, 4, 5) of AB. ∴ M= ⎛ 1 + 3 , 2+ 4, 3+ 5⎞ = (2, 3, 4) ⎜⎝ 2 2 2 ⎠⎟ Let P(x, y, z) be a general point on the plane. ∴ d.r.’s of MP are x – 2, y – 3, z – 4 and d.r.’s of AB are 3 – 1, 4 – 2, 5 – 3 or 2, 2, 2 or 1, 1, 1. Since AB ⊥ MP, we have (1)(x – 2) + (1)(y – 3) + 1(z – 4) = 0. ⇒ x – 2 + y – 3 + z – 4 = 0 ⇒ x + y + z = 9. This is the equation of the required plane. Express the straight line x + y + z + 1 = 0, 4x + y – 2z + 2 = 0 in the symmmetric form. Given equations are x+y+z+1=0 ...(1) 4x + y – 2z + 2 = 0. ...(2) We find two equations each involving only two variables. (1) – (2) ⇒ – 3x + 3z – 1 = 0 ...(3) 2 × (1) + (2) ⇒ 6x + 3y + 4 = 0 ...(4) (3) and (4) are the desired equations. The variable x is involved in both the equations. (3) ⇒ x= 3z − 1 and (4) ⇒ x=– 3y + 4 3 6 x=– 3y + 4 3z − 1 x= y + 4 z− 1 6 3 3 1 3 ⇒ = ⇒ −2 = x − 0 y + 4 z − 1 . 1 3 1 3 ⇒ = −2 = This is the symmetric form of the given line.

272 Mathematics—XII Example 9. Find the equation of the plane passing through the points (–2, 6, –6), (–3, 10, –9) Solution. and (–5, 0, –6). Let the given points be A(–2, 6, –6), B(–3, 10, –9) and C(–5, 0, –6). Let →r be the position vector of a general point P(x, y, z) on the plane. → Now AP = P.V. of P – P.V of A = (xiˆ + yˆj + zkˆ) − (− 2iˆ + 6ˆj − 6kˆ) = (x + 2) iˆ + (y – 6) ˆj + (z + 6) kˆ , → AB = P.V. of B – P.V. of A = (−3iˆ + 10ˆj – 9kˆ) − (−2iˆ + 6ˆj − 6kˆ) = − iˆ + 4ˆj − 3kˆ → and AC = P.V. of C – P.V. of A = – 5iˆ + 0ˆj − 6kˆ − (−2iˆ + 6ˆj − 6kˆ) = −3iˆ − 6ˆj →→ → Since P is on the plane passing through A, B, C, the vectors AP, AB and AC are coplanar. ∴ [ → → → ] = 0 AP AB AC x+2 y−6 z+6 ⇒ −1 4 −3 = 0 −3 −6 0 ⇒ (x + 2)(0 – 18) – (y – 6)(0 – 9) + (z + 6)(6 + 12) = 0 ⇒ 2x – y – 2z – 2 = 0. This is the required equation. Example 10. Find the equation of the plane passing through the points (–2, 6, –6), (–3, 10, –9) and (–5, 0, –6). Solution. Let the given points be A(–2, 6, –6), B(–3, 10, –9) and C(–5, 0, –6). Let the equation of the required plane be a(x + 2) + b(y – 6) + c(z + 6) = 0. ...(1) B(–3, 10, –9) is on the plane (1). (By using point A) ⇒ a(–3 + 2) + b(10 – 6) + c(–9 + 6) = 0 ⇒ – a + 4b – 3c = 0 ...(2) C(–5, 0, –6) is also on the plane (1). ⇒ – 3a – 6b + 0c = 0 ...(3) ⇒ a(–5 + 2) + b(0 – 6) + c(–6 + 6) = 0 (2) and (3) imply 0 a = 9 b 0 = 6 c or a = b = c = k (say) − 18 + + 12 −2 1 2 ⇒ a = –2k, b = k, c = 2k ∴ (1) ⇒ –2k(x + 2) + k(y – 6) + 2k(z + 6) = 0 ⇒ –2x – 4 + y – 6 + 2z + 12 = 0 ⇒ 2x – y – 2z – 2 = 0. This is the required equation.

Three Dimensional Geometry 273 Alternatively, eliminating a, b, c from (1), (2), (3), we get x+2 y−6 z+6 = 0. −1 4 −3 −3 0 −6 ⇒ (x + 2)(0 – 18) – (y – 6)(0 – 9) + (z + 6)(6 + 12) = 0 ⇒ –18x – 36 + 9y – 54 + 18z + 108 = 0 ⇒ 18x – 9y – 18z – 18 = 0 ⇒ 2x – y – 2z – 2 = 0. Example 11. Find the equation of the plane with intercept 3 on the y-axis and parallel to ZOX plane. (NCERT) Solution. Let the equation of the plane be x + y + z = 1. ...(1) a b c Here b = 3 ∴ (1) ⇒ x + y + z =1 ...(2) a 3 c ∴ d.r.’s of normal to the plane (2) are 1 , 1 , 1 . a 3 c This plane is parallel to ZOX plane. ∴ y-axis is normal to the plane (2). Also, d.r.’s of the y-axis are 0, 1, 0. ∴ 1/a = 1/3 = 1/c 0 1 0 ⇒ 1 = 0 and 1 =0 a c ∴ (2) ⇒ 0⋅x+ y + 0 ⋅ z = 1, i.e., y – 3 = 0. 3 This is the equation of the required plane. Example 12. A plane meets the coordinate axes in A, B and C such that the centroid of ΔABC is the point (α, β, γ). Show that the equation of the plane is x + y + z = 3. α β γ [NCERT (EP); CBSE 2018 SP] Solution. Let the equation of the plane be x + y + z = 1. ...(1) a b c ∴ Intercepts on the axes are a, b and c. ∴ The coordinates of A, B and C are (a, 0, 0), (0, b, 0) and (0, 0, c) respectively. The centroid of ΔABC is (α, β, γ). ∴ a+0+0 = α, 0+b+0 =β and 0+0+c =γ 3 3 3 ∴ a = 3α, b = 3β and c = 3γ ∴ (1) ⇒ x + y + z =1 ⇒ x y z = 3. 3α 3β 3γ ++ αβγ This is the required equation.

274 Mathematics—XII Example 13. Find the equation of the line passing through the point with position vector iˆ + ˆj + kˆ and perpendicular to the plane (2iˆ + ˆj + 3kˆ) ⋅ →r = 5. Solution. The given plane is (2iˆ + ˆj + 3kˆ) ⋅ →r = 5. ⇒ (2iˆ + ˆj + 3kˆ) ⋅ (xiˆ + yˆj + zkˆ) = 5 ⇒ 2x + y + 3z = 5 ∴ 2, 1, 3 are d.r.’s of normal to the plane. ∴ The required line has d.r.’s 2, 1, 3. Also, the line passes through the point whose position vector is iˆ + ˆj + kˆ . ∴ The line passes through (1, 1, 1). ∴ The equations of the required line are x−1 = y − 1 = z − 1 . 2 1 3 Example 14. Find the equation of the plane passing through the point (1, 1, 1) and perpendicular to the line x – 2y + z = 2, 4x + 3y – z + 1 = 0. Solution. The equations of the line are x – 2y + z = 2 ...(1) and 4x + 3y – z + 1 = 0. ...(2) This line is the intersection of the planes (1) and (2). Let a, b, c be d.r.’s of the given line. The given line is perpendicular to the normals to the planes (1) and (2). 1, –2, 1 and 4, 3, –1 are d.r.’s of normals to the planes (1) and (2) respectively. ∴ 1 ⋅ a – 2b + 1 ⋅ c = 0 ...(3) and 4a + 3b – 1 ⋅ c = 0 ...(4) Solving (3) and (4), we get a = 4 b 1 = c or a = b = c . 2−3 + 3+8 −1 5 11 ∴ –1, 5, 11 are d.r.’s of the given line, because a, b, c are proportional to –1, 5, 11. The required plane is perpendicular to the given line. ∴ –1, 5, 11 are d.r.’s of normal to the required plane. Let the equation of the required plane be –1 . x + 5y + 11z = d. (1, 1, 1) lies on this plane. ∴ –1(1) + 5(1) + 11(1) = d or d = 15. ∴ The equation of the required plane is –x + 5y + 11z = 15. Example 15. Find the distance of the point (–2, 3, –4) from the line x+2 = 2y + 3 = 3z + 4 3 4 5 measured parallel to the plane 4x + 12y – 3z + 1 = 0. Solution. Let the given point be P(–2, 3, –4). Given line is x+2 = 2y + 3 = 3z + 4 = λ, say. 3 4 5 Let Q ⎛ 3λ − 2, 4λ − 3, 5λ − 4⎞ be any point on this line. ⎜⎝ 2 3 ⎠⎟

Three Dimensional Geometry 275 ∴ d.r.’s of PQ are (3λ – 2) – (–2), 4λ − 3 – 3, 5λ − 4 – (–4) 2 3 i.e., 3λ, 4λ − 9 , 5λ + 8 . 2 3 Let PQ be parallel to the plane 4x + 12y – 3z + 1 = 0. d.r.’s of normal to the plane are 4, 12, –3. ∴ 4(3λ) + 12 ⎛ 4λ − 9 ⎞ – 3 ⎛ 5λ + 8 ⎞ =0 ⎝⎜ 2 ⎠⎟ ⎝⎜ 3 ⎟⎠ ⇒ 12λ + 24λ – 54 – 5λ – 8 = 0 ⇒ 31λ = 62 ⇒ λ = 2 ∴ Coordinates of Q are ⎛ 3(2) − 2, 4(2) − 3 , 5(2) − 4 ⎞ i.e., ⎛ 4, 5 , 2 ⎞ . ⎝⎜ 2 3 ⎠⎟ ⎝⎜ 2 ⎠⎟ ∴ Required distance, PQ = (4 + 2)2 + ⎛ 5 − 3 ⎞2 + (2 + 4)2 ⎜⎝ 2 ⎠⎟ = 36 + 1 + 36 = 289 = 17 units. 4 4 2 Example 16. Find the coordinates of the foot of perpendicular drawn from the origin to the plane 2x – 3y + 4z – 6 = 0. (NCERT) Solution. The given plane is 2x – 3y + 4z – 6 = 0. ...(1) d.r.’s of normal are 2, –3, 4. Let P(α, β, γ) be the foot of perpendicular from O. ∴ d.r.’s of OP are α – 0, β – 0, γ – 0, O (0, 0, 0) i.e., α, β, γ. OP and normal to plane are parallel. ∴ α = β = γ =λ (say) 2 −3 4 P(a, b, g) ∴ α = 2λ, β = –3λ, γ = 4λ Since P is on the plane, we have 2α – 3β + 4γ – 6 = 0. ⇒ 2(2λ) – 3(–3λ) + 4(4λ) – 6 = 0 ⇒ λ = 6/29 ∴ α= 2 ⎛ 6 ⎞ = 12 , β = −3 ⎛ 6 ⎞ = − 18 , γ = 4 ⎛ 6 ⎞ = 24 ⎝⎜ 29 ⎟⎠ 29 ⎝⎜ 29 ⎠⎟ 29 ⎜⎝ 29 ⎟⎠ 29 ∴ Foot of perpendicular, P = ⎛ 12 , − 18 , 24 ⎞ . ⎜⎝ 29 29 29 ⎠⎟ Example 17. Find the image of the point (1, 3, 4) in the plane x – y + z = 5. Solution. The given plane is x – y + z = 5. ...(1) d.r.’s of normal are 1, –1, 1. Let B(α, β, γ) be the image of A. ∴ d.r.’s of AB are α – 1, β – 3, γ – 4.

276 Mathematics—XII AB and normal to plane are parallel. ∴ α − 1 = β−3 = γ −4 = λ (say) 1 −1 1 ∴ α = λ + 1, β = –λ + 3, γ = λ + 4 ...(2) Let M be the midpoint of AB. ∴ M = ⎛ α+ 1, β + 3 , γ + 4 ⎞ ⎜⎝ 2 2 2 ⎠⎟ = ⎛ λ +1 + 1 , −λ +3 + 3 , λ + 4 + 4 ⎞ ⎝⎜ 2 2 2 ⎟⎠ A(1, 3, 4) = ⎛ λ + 2 , 6 − λ , λ + 8 ⎞ ⎝⎜ 2 2 2 ⎠⎟ Since M is on the plane, we have ⎛ λ + 2 ⎞ − ⎛ 6 − λ ⎞ + ⎛ λ + 8 ⎞ = 5. ⎝⎜ 2 ⎠⎟ ⎜⎝ 2 ⎠⎟ ⎜⎝ 2 ⎠⎟ M ⇒ λ + 2 – 6 + λ + λ + 8 = 10 ⇒ 3λ = 6 ⇒ λ = 2 ∴ (2) ⇒ α = 2 + 1 = 3, β = –2 + 3 = 1, γ = 2 + 4 = 6 B(α, β, γ) ∴ Image of the given point is (3, 1, 6). Example 18. Find the vector equation of the plane passing through three points with position vectors iˆ + ˆj – 2kˆ , 2iˆ – ˆj + kˆ and iˆ + 2ˆj + kˆ . Also, find the coordinates of the Solution. point of intersection of this plane and the line →r = 3iˆ – ˆj – kˆ + λ(2iˆ – 2ˆj + kˆ) . (CBSE 2013) Let A, B, C be points such that P.V. of A = iˆ + ˆj – 2kˆ , P.V. of B = 2iˆ – ˆj + kˆ , P.V. of C = iˆ + 2ˆj + kˆ . Let →r be the position vector of a general point P(x, y, z) on the plane through A, B and C. Now, A→P = P.V. of P – P.V. of A = (xiˆ + yˆj + zkˆ) – (iˆ + ˆj – 2kˆ) = (x – 1)iˆ + (y – 1)ˆj + (z + 2)kˆ A→B = P.V. of B – P.V. of A = (2iˆ – ˆj + kˆ) – (iˆ + ˆj – 2kˆ) = iˆ – 2ˆj + 3kˆ A→C = P.V. of C – P.V. of A = (iˆ + 2ˆj + kˆ) – (iˆ + ˆj – 2kˆ) = ˆj + 3kˆ The vectors A→P , A→B and A→C are coplanar. ∴ [ A→P A→B A→C ] = 0 x–1 y–1 z+2 ⇒ 1 –2 3 = 0 013

Three Dimensional Geometry 277 ⇒ (x – 1)(–6 – 3) – (y – 1)(3 – 0) + (z + 2)(1 – 0) = 0 ...(1) ⇒ –9x – 3y + z + 14 = 0 ⇒ 9x + 3y – z = 14 ⇒ (9iˆ + 3 jˆ – kˆ) ⋅ →r = 14. This is the equation of the required plane. The given line is →r = 3iˆ – ˆj – kˆ + λ(2iˆ – 2 ˆj + kˆ) . ...(2) Solving (1) and (2), we get (9iˆ + 3ˆj – kˆ) ⋅ [3iˆ – ˆj – kˆ + λ(2iˆ – 2ˆj + kˆ)] = 14. ⇒ (9iˆ + 3ˆj – kˆ) ⋅ [(3 + 2λ)iˆ + (–1 – 2λ)ˆj + (–1 + λ)kˆ] = 14 ⇒ 9(3 + 2λ) + 3(–1 – 2λ) –1(–1 + λ) = 14 ⇒ 27 + 18λ – 3 – 6λ + 1 – λ = 14 ⇒ 11λ = –11 ⇒ λ = –11/11 = –1 ∴ (2) ⇒ →r = 3iˆ – ˆj – kˆ + (–1)(2iˆ – 2 ˆj + kˆ) = iˆ + ˆj – 2kˆ = P.V. of (1, 1, –2) ∴ The point of intersection is (1, 1, –2). Example 19. Find the coordinates of the point where the line through (3, –4, –5) and (2, –3, 1) crosses the plane through the points (2, 2, 1), (3, 0, 1) and (4, –1, 0). [NCERT (EP); A.I. CBSE 2013] Solution. Using two-point form, the equations of the line passing through (3, –4, –5) and (2, –3, 1) are x − 3 = y – (–4) = z – (–5) . 2 – 3 –3 – (–4) 1 – (–5) ⇒ x−3 = y + 4 = z + 5 ...(1) –1 1 6 ...(2) ...(3) Let the plane passing through (2, 2, 1), (3, 0, 1) and (4, –1, 0) be ...(4) ...(5) ax + by + cz = d. ...(6) (2, 2, 1) is on (2). ...(7) ∴ 2a + 2b + c = d (2) – (3) ⇒ a(x – 2) + b(y – 2) + c(z – 1) = 0 (3, 0, 1) is on the plane. (4) ⇒ a(3 – 2) + b(0 – 2) + c(1 – 1) = 0 ⇒ a – 2b + 0.c = 0 (4, –1, 0) is on the plane. ∴ (4) ⇒ a(4 – 2) + b(–1 – 2) + c(0 – 1) = 0 ⇒ 2a – 3b – c = 0 (5) and (6) imply a = 0 b 1 = c 4 or a = b = c = k, say 2–0 + –3 + 2 1 1 ⇒ a = 2k, b = k, c = k ∴ (4) ⇒ 2k(x – 2) + k(y – 2) + k(z – 1) = 0 ⇒ 2x – 4 + y – 2 + z – 1 = 0 ⇒ 2x + y + z = 7 Any point on the line (1) is (–λ + 3, λ – 4, 6λ – 5). Let this point be on the plane (7).

278 Mathematics—XII ∴ 2(–λ + 3) + (λ – 4) + (6λ – 5) = 7 ⇒ 5λ = 10 ⇒ λ = 2 ∴ The point of intersection is (–2 + 3, 2 – 4, 6(2) – 5) i.e., (1, –2, 7). Example 20. Find the distance of the point (3, 4, 5) from the plane x + y + z = 2 measured parallel to the line 2x = y = z. (CBSE 2012 C) Solution. The given line is 2x = y = z. ⇒ x = y = z ⇒ x = y = z . 1/2 1 1 1 2 2 ∴ 1, 2, 2 are d.r.’s of the given line. ∴ The line passing through A(3, 4, 5) and parallel to the given line is x – 3 = y – 4 = z – 5 = λ (say) ...(1) 1 2 2 General point on (1) is B(λ + 3, 2λ + 4, 2λ + 5). Let this point lie on the given plane x + y + z = 2. ⇒ (λ + 3) + (2λ + 4) + (2λ + 5) = 2 ⇒ 5λ = –10 ⇒ λ = –2 ∴ λ + 3 = –2 + 3 = 1, 2λ + 4 = 2(–2) + 4 = 0, 2λ + 5 = 2(–2) + 5 = 1 ∴ Coordinates of B are (1, 0, 1). ∴ Required distance AB = (3 – 1)2 + (4 – 0)2 + (5 – 1)2 = 4 + 16 + 16 = 6 units. Example 21. Find the equation of the plane containing the coplanar lines: →r = iˆ + ˆj + kˆ + λ(iˆ − ˆj + kˆ) and →r = 4ˆj + 2kˆ + μ(2iˆ − ˆj + 3kˆ) . Solution. Given lines are → = iˆ + ˆj + kˆ + λ(iˆ − ˆj + kˆ) ...(1) r and →r = 4ˆj + 2kˆ + μ(2iˆ − ˆj + 3kˆ) . ...(2) (1) ⇒ →r = →a + λ→b , where →a = iˆ + ˆj + kˆ and →b = iˆ − ˆj + kˆ (2) ⇒ →r = →c + μ→d , where →c = 4 ˆj + 2kˆ and →d = 2iˆ − ˆj + 3kˆ Let P(x, y, z) be a general point on the required plane and let →r be its position vector. Let →a be the position vector of the point A(1, 1, 1). ∴ → = P.V. of P – P.V. of A = →r − →a = (x − 1)iˆ + (y − 1)ˆj + (z − 1)kˆ AP The vectors →→ and → are coplanar. ∴ → →→ AP, b [AP b d ] = 0 d x−1 y−1 z−1 ∴ 1 −1 1 = 0 2 −1 3 ⇒ (x – 1)(–3 + 1) – (y – 1)(3 – 2) + (z – 1)(–1 + 2) = 0 ⇒ –2x + 2 – y + 1 + z – 1 = 0 ⇒ 2x + y – z = 2. This is the equation of the required plane.

Three Dimensional Geometry 279 Example 22. Show that the lines x + 4 = y + 6 = z−1 and 3x – 2y + z + 5 = 0 = 2x + 3y + 3 5 −2 4z – 4 intersect. Find the equation of the plane in which they lie and also find their point of intersection. Solution. The given lines are x + 4 = y + 6 = z−1 ...(1) 3 5 −2 and 3x – 2y + z + 5 = 0 = 2x + 3y + 4z – 4. ...(2) Line (1) passes through the point A(–4, –6, 1) and has d.r.’s 3, 5, –2. Line (2) is the line of intersection of the planes 3x – 2y + z + 5 = 0 ...(3) and 2x + 3y + 4z – 4 = 0. ...(4) Let a, b, c be d.r.’s of the line (2). The line (2) is perpendicular to normals to planes (3) and (4). ∴ 3a – 2b + c = 0 and 2a + 3b + 4c = 0 ⇒ a 3 = 2 b = 9 c 4 ⇒ a = b = c ⇒ a = b = c −8 − − 12 + −11 −10 13 11 10 −13 ∴ d.r.’s of line (2) are 11, 10, –13. Let lines (1) and (2) intersect at the point B(3λ – 4, 5λ – 6, –2λ + 1), which is always on the line (1), for all values of λ. Since B is on the line (2), so it lies on planes (3) and (4) both. ∴ 3(3λ – 4) – 2(5λ – 6) + (–2λ + 1) + 5 = 0 ...(5) and 2(3λ – 4) + 3(5λ – 6) + 4(–2λ + 1) – 4 = 0 ...(6) (5) ⇒ – 3λ + 6 = 0 ⇒ λ = 2 and (6) ⇒ 13λ – 26 = 0 ⇒ λ = 2 ∴ The line (1) meets the planes (3) and (4) at the same point. ∴ The lines (1) and (2) intersect at B. Putting λ = 2 in the coordinates of B, we get B = (3(2) – 4, 5(2) – 6, –2(2) + 1) = (2, 4, –3). Since (2, 4, –3) is also on the line (2), the equations of this line can be written as x−2 = y−4 = z+3 . ...(7) 11 10 −13 The equation of the plane containing lines (1) and (7) is x+4 y+6 z−1 3 5 −2 = 0 . 11 10 −13 ⇒ (x + 4)(–65 + 20) – (y + 6)(–39 + 22) + (z – 1)(30 – 55) = 0 ⇒ – 45x – 180 + 17y + 102 – 25z + 25 = 0 ⇒ 45x – 17y + 25z + 53 = 0. This is the equation of the required plane.

280 Mathematics—XII Example 23. Find the equation of the plane passing through the points (3, 4, 1) and (0, 1, 0) and parallel to the line x + 3 = y − 3 = z − 2 . 2 7 5 Solution. Given line is x + 3 = y − 3 = z − 2 . ...(1) 2 7 5 Line (1) passes through (–3, 3, 2) and has d.r.’s 2, 7, 5. Let the equation of the required plane be a(x – 3) + b(y – 4) + c(z – 1) = 0. ...(2) (∵ (3, 4, 1) is on the required plane) (0, 1, 0) is on the plane (2). ∴ a(0 – 3) + b(1 – 4) + c(0 – 1) = 0 ⇒ – 3a – 3b – c = 0 ...(3) The line (1) is perpendicular to the normal to (2). ∴ 2a + 7b + 5c = 0 ...(4) Solving (3) and (4), we get a = b = c 6 or a = b = c = λ, say −15 + 7 −2 + 15 −21 + −8 13 −15 ∴ a = –8λ, b = 13λ, c = –15λ Putting the values of a, b, c in (2), we get –8λ(x – 3) + 13λ(y – 4) – 15λ(z – 1) = 0. ⇒ – 8x + 24 + 13y – 52 – 15z + 15 = 0 ⇒ 8x – 13y + 15z + 13 = 0. This is the equation of the required plane. Alternatively, eliminating a, b, c from (2), (3), (4), we get x−3 y−4 z−1 −3 −3 −1 = 0. 2 75 ⇒ (x – 3)(–15 + 7) – (y – 4)(–15 + 2) + (z – 1)(–21 + 6) = 0 ⇒ – 8(x – 3) + 13(y – 4) – 15(z – 1) = 0 ⇒ 8x – 13y + 15z + 13 = 0. Example 24. Find the equation of the plane which contains the line →r = 2iˆ + λ(ˆj − kˆ) and perpendicular to the plane (iˆ + kˆ) ⋅ →r = 3 . Find the position vector of the point, where this plane meets the line → = λ(2iˆ + 3ˆj + kˆ) . r Solution. The given line and the plane are respectively → = 2iˆ + λ( ˆj − kˆ) ...(1) and (iˆ + kˆ) ⋅ → = 3 . ...(2) r r The line (1) passes through the point (2, 0, 0) and has d.r.’s 0, 1, –1. Since, the required plane passes through (2, 0, 0), so let its equation be a(x – 2) + b(y – 0) + c(z – 0) = 0. ...(3) a, b, c are d.r.’s of normal to plane (3). ∴ 0⋅a + 1⋅b + (–1)⋅c = 0 (∵ Line (1) is ⊥ to normal to (3)) The plane (3) is perpendicular to plane (2).

Three Dimensional Geometry 281 ∴ 1⋅a + 0⋅b + 1⋅c = 0 (∵ 1, 0, 1 are d.r.’s of normal to (2)) Solving, we get a = b = c or a = b = c = λ, say 1−0 −1 − 0 0−1 1 −1 −1 ∴ a = λ, b = –λ, c = –λ Putting these values in (3), we get λ(x – 2) + (–λ)y + (–λ)z = 0. ⇒ x – 2 – y – z = 0 or x – y – z = 2. ...(4) The given straight line, →r = λ(2iˆ + 3 ˆj + kˆ) is equivalent to x − 0 = y − 0 = z −0 = λ, where →r = xiˆ + yˆj + zkˆ . ...(5) 2 3 1 Let the line (5) meet the plane (4) at the point (2λ, 3λ, λ). ∴ (4) ⇒ 2λ – 3λ – λ = 2 ⇒ λ = –1 ∴ The point of intersection is (2(–1), 3(–1), –1) = (–2, –3, –1). The P.V. of this point is – 2iˆ – 3 jˆ – kˆ. Example 25. Find the equation of the plane passing through the point (1, 2, 5) and parallel to the lines x − 1 = y+1 = z and x+3 = y− 3 = z + 1 . 2 3 1 −2 5 3 Solution. Let the equation of the required plane be a(x – 1) + b(y – 2) + c(z – 5) = 0. ...(1) (∵ (1, 2, 5) is on the required plane) The normal to the plane is perpendicular to the lines x − 1 = y + 1 = z 2 3 1 and x+3 = y − 3 = z + 1 and their d.r.’s are 2, 3, 1 and –2, 5, 3 respectively. −2 5 3 ∴ 2a + 3b + 1c = 0 and –2a + 5b + 3c = 0 Solving these, we get 9 a 5 = b 6 = c 6 or a = 6 = c or a = 6 = c = λ, say − −2 − 10 + 4 −8 16 1 −2 4 ∴ a = λ, b = –2λ, c = 4λ Putting the values of a, b, c in (1), we get λ(x – 1) – 2λ(y – 2) + 4λ(z – 5) = 0. ⇒ x – 1 – 2y + 4 + 4z – 20 = 0 ⇒ x – 2y + 4z = 17. This is the equation of the required plane. Example 26. The equation of a plane is → = iˆ − ˆj + λ(iˆ + ˆj + kˆ) + μ(4iˆ − 2ˆj + 3kˆ) , where λ r and μ are arbitrary scalars. Express the equation of this plane in the scalar product form. Solution. The plane is →r = iˆ − ˆj + λ(iˆ + ˆj + kˆ) + μ(4iˆ − 2ˆj + 3kˆ) . ...(1) Let →r = xiˆ + yˆj + zkˆ , where (x, y, z) is a general point on the plane (1). ∴ (1) ⇒ xiˆ + yˆj + zkˆ = iˆ − ˆj + λ ( iˆ + ˆj + kˆ ) + μ( 4iˆ − 2ˆj + 3kˆ )

282 Mathematics—XII ⇒ xiˆ + yˆj + zkˆ = (1 + λ + 4μ) iˆ + (−1 + λ − 2μ)ˆj + (λ + 3μ)kˆ ⇒ 1 + λ + 4μ = x ...(2) –1 + λ – 2μ = y ...(3) λ + 3μ = z ...(4) Solving (2) and (3), we get λ= x + 2y + 1 and μ= x − y − 2 . 3 6 ∴ (4) ⇒ x + 2y + 1 + 3 ⎛ x − y − 2 ⎞ =z ⇒ 2x + 4y + 2 + 3x – 3y – 6 = 6z 3 ⎝⎜ 6 ⎟⎠ ⇒ 5x + y – 6z = 4 ⇒ (5iˆ + jˆ – 6kˆ ) . → = 4. r This is the equation of the given plane in the scalar product form. Example 27. Find the distance of the point (2, –1, 3) from the plane Solution. (5iˆ + 2ˆj − 7kˆ) ⋅ →r + 12 = 0. Given plane is →n ⋅ →r = d, where →n = 5iˆ + 2 ˆj − 7kˆ and d = –12. Let →a be the position vector of (2, –1, 3). ∴ → = 2iˆ − ˆj + 3kˆ a Distance of (2, –1, 3) from the plane = →n ⋅ →a − d = (5iˆ + 2ˆj − 7kˆ) ⋅ (2iˆ − ˆj + 3kˆ) + 12 |→n| 52 + 22 + (−7)2 = 10 − 2 − 21 + 12 = 1 unit. 78 78 Example 28. Find the distance between the parallel planes 2x – y + 3z = 4 and 2x – y + 3z = 18. (CBSE 2012 SP) Solution. Given planes are 2x – y + 3z – 4 = 0 ...(1) and 2x – y + 3z – 18 = 0. ...(2) Let P(α, β, γ) be any point on (1). ∴ 2α – β + 3γ – 4 = 0 ...(3) Distance between planes = Distance of P(α, β, γ) from plane (2) = 2α – β + 3γ – 18 = 4 – 18 = 14 = 14 units. (Using (3)) 4 +1+ 9 14 14 Example 29. Show that the points iˆ − ˆj + 3kˆ and 3 (iˆ + ˆj + kˆ) are equidistant from the plane (5iˆ + 2ˆj − 7kˆ) ⋅ → + 9 = 0 and are on the opposite sides of it. [NCERT (EP)] r Solution. The given plane is (5iˆ + 2 ˆj − 7 kˆ ) ⋅ → + 9 = 0. r ⇒ 5x + 2y – 7z + 9 = 0, where → = xiˆ + yˆj + zkˆ . r

Three Dimensional Geometry 283 Let iˆ − ˆj + 3kˆ and 3 (iˆ + ˆj + kˆ) be the P.V. of points A and B. ∴ A = (1, –1, 3) and B = (3, 3, 3) ∴ Distance of A from the plane = 5(1) + 2(−1) − 7(3) + 9 = 5 − 2 − 21 + 9 = 9 units 25 + 4 + 49 78 78 Distance of B from the plane = 5(3) + 2(3) − 7(3) + 9 = 15 + 6 − 21 + 9 = 9 units 25 + 4 + 49 78 78 ∴ The points A and B are equidistant from the plane. The points A and B lies on the opposite sides, because 5(1) + 2(–1) – 7(3) + 9 = –9 and 5(3) + 2(3) – 7(3) + 9 = 9 are of opposite signs. Example 30. Find the equation of the plane through the points A(1, 1, 0), B(1, 2, 1) and C(–2, 2, –1) and hence find the distance between the plane and the line x – 6 = y–3 = z + 2 . (CBSE 2014 SP) 3 –1 1 Solution. Let the equation of the plane through A, B and C be ax + by + cz = d. ...(1) A(1, 1, 0) is on (1). ∴ a(1) + b(1) + c(0) = d ...(2) (1) – (2) ⇒ a(x – 1) + b(y – 1) + cz = 0 ...(3) B(1, 2, 1) is on the plane. ∴ (3) ⇒ a(1 – 1) + b(2 – 1) + c(1) = 0 ⇒ 0⋅a + b + c = 0 ...(4) C(–2, 2, –1) is on the plane. ∴ (3) ⇒ a(–2 – 1) + b(2 – 1) + c(–1) = 0 ⇒ –3a + b – c = 0 ...(5) (4) and (5) imply a 1 = b 0 = c or a = b = c = k, say –1 – –3 – 0+3 –2 –3 3 ⇒ a = –2k, b = –3k, c = 3k ...(6) ∴ (3) ⇒ –2k(x – 1) – 3k(y – 1) + 3kz = 0 ⇒ –2x + 2 – 3y + 3 + 3z = 0 ⇒ 2x + 3y – 3z – 5 = 0. This is the equation of the required plane. The given line is x – 6 = y–3 = z+ 2 . ...(7) 3 –1 1 d.r. of normal to plane (6) are 2, 3, –3 and d.r. of line (7) are 3, –1, 1. Here 2(3) + 3(–1) + (–3)(1) = 6 – 3 – 3 = 0. ∴ The line (7) is parallel to the plane (6). The point (6, 3, –2) is on the line. ∴ Distance between plane and line = Perpendicular distance of (6, 3, –2) from the plane = 2(6) + 3(3) – 3(–2) – 5 = 12 + 9 + 6 – 5 = 22 = 22 units 4+9+9 22 22

284 Mathematics—XII Example 31. Find the equation of the plane passing through the point (2, 1, –1) and the intersection of the planes (iˆ + 3ˆj − kˆ) ⋅ →r = 0 and (ˆj + 2kˆ) ⋅ →r = 0. Solution. Given planes are (iˆ + 3 ˆj − kˆ) ⋅ →r = 0 ...(1) and (ˆj + 2kˆ) ⋅ →r = 0. ...(2) (1) ⇒ x + 3y – z = 0 and (2) ⇒ y + 2z = 0 Let the equation of the required plane be x + 3y – z + λ(y + 2z) = 0. The point (2, 1, –1) is on this plane. ∴ 2 + 3(1) + 1 + λ(1 + 2(–1)) = 0 ⇒ 6 – λ = 0 ⇒ λ = 6 ∴ The required plane is x + 3y – z + 6(y + 2z) = 0 or x + 9y + 11z = 0. Example 32. Find the equation of the plane containing the line of intersection of the planes x + 2y + 3z – 4 = 0, 2x + y – z + 5 = 0 and perpendicular to the plane 5x + 3y + 6z + 8 = 0. [NCERT (EP)] Solution. Given planes are x + 2y + 3z – 4 = 0 ...(1) and 2x + y – z + 5 = 0. ...(2) Let the equation of the required plane be x + 2y + 3z – 4 + λ(2x + y – z + 5) = 0. ...(3) ⇒ (1 + 2λ)x + (2 + λ)y + (3 – λ)z – 4 + 5λ = 0 This plane is perpendicular to the plane 5x + 3y + 6z + 8 = 0. ∴ 5(1 + 2λ) + 3(2 + λ) + 6(3 – λ) = 0 ⇒ 5 + 10λ + 6 + 3λ + 18 – 6λ = 0 ⇒ 7λ + 29 = 0 ⇒ λ = –29/7 ∴ (3) ⇒ x + 2y + 3z – 4 – 29 (2x + y – z + 5) = 0 7 ⇒ 7x + 14y + 21z – 28 – 58x – 29y + 29z – 145 = 0 ⇒ – 51x – 15y + 50z – 173 = 0 ⇒ 51x + 15y – 50z + 173 = 0. This is the required plane. Example 33. Find the equation of the plane through the line of intersection of → . (2iˆ − 3 ˆj + 4kˆ) r = 1 and →r . (iˆ − ˆj) + 4 = 0 and perpendicular to the plane →r . (2iˆ − ˆj + kˆ) + 8 = 0. Hence, find whether the plane thus obtained contains the line x – 1 = 2y – 4 = 3z – 12. (CBSE 2017) Solution. Given planes are →r . (2iˆ − 3ˆj + 4kˆ) = 1 ...(1) and →r . (iˆ − ˆj) + 4 = 0. ...(2) ...(4) (1) ⇒ 2x – 3y + 4z – 1 = 0 ...(3) and (2) ⇒ x – y + 4 = 0 Let the equation of the required plane be ...(5) ...(6) 2x – 3y + 4z – 1 + λ(x – y + 4) = 0. ⇒ (2 + λ)x – (3 + λ)y + 4z – 1 + 4λ = 0 (6) is perpendicular to →r . (2iˆ − ˆj + kˆ) + 8 = 0 i.e., 2x – y + z + 8 = 0. ⇒ 2(2 + λ) + 1(3 + λ) + 1(4) = 0 284

Three Dimensional Geometry 285 ⇒ 4 + 2λ + 3 + λ + 4 = 0 ⇒ 3λ + 11 = 0 ⇒ λ = –11/3 ∴ (5) ⇒ 2x – 3y + 4z – 1 + (–11/3)(x – y + 4) = 0 ⇒ 6x – 9y + 12z – 3 – 11x + 11y – 44 = 0 ⇒ 5x – 2y – 12z + 47 = 0. This is the required equation. Given line is x – 1 = 2y – 4 = 3z – 12. ⇒ x–1= y−2 = z−4 ⇒ x−1 = y−2 = z−4 ...(7) 1/2 1/3 6 3 2 This line is parallel to the plane 5x – 2y – 12z + 47 = 0 if 6(5) + 3(–2) + 2 (–12) = 0 or if 30 – 6 – 24 = 0 or if 0 = 0, which is true. ∴ Given line and plane are parallel. Line (7) passes through (1, 2, 4) and this point lies on the plane 5x – 2y – 12z + 47 = 0 if 5(1) – 2(2) – 12(4) + 47 = 0 or if 5 – 4 – 48 + 47 = 0 or if 0 = 0, which is true. ∴ The point (1, 2, 4) lies on the plane 5x – 2y – 12z + 47 = 0. ∴ The plane 5x – 2y – 12z + 47 = 0 contains the given line. Example 34. Find the equation of the plane through the line of intersection of the planes x + y + z = 1 and 2x + 3y + 4z = 5 and twice of its y-intercept is equal to three times its z-intercept. (CBSE 2017 C) Solution. Given planes are x+y+z–1=0 ...(1) and 2x + 3y + 4z – 5 = 0. ...(2) Let the equation of the required plane be x + y + z – 1 + λ(2x + 3y + 4z – 5) = 0. ...(3) ⇒ (1 + 2λ)x + (1 + 3λ)y + (1 + 4λ)z = 1 + 5λ ⇒ 1 x + 1 y + z =1 + 5λ + 5λ 1 + 5λ 1 + 2λ 1 + 3λ 1 + 4λ ∴ y-intercept = 1+ 5λ and z-intercept = 1+ 5λ 1+ 3λ 1+ 4λ By the given condition, 2 ⎛ 1 + 5λ ⎞ = 3 ⎛ 1 + 5λ ⎞ . ⎜ 1 + ⎟ ⎜ 1 + 4λ ⎟ ⎝ 3λ ⎠ ⎝ ⎠ ⇒ (1 + 5λ)[2(1 + 4λ) – 3(1 + 3λ)] = 0 ⇒ (1 + 5λ)(–λ – 1) = 0 ⇒ λ = –1/5, –1 Case I λ = –1/5 (3) ⇒ x + y + z – 1 + (–1/5)(2x + 3y + 4z – 5) = 0 ⇒ 5x + 5y + 5z – 5 – 2x – 3y – 4z + 5 = 0 ⇒ 3x + 2y + z = 0 This plane passes through the origin and does not satisfy the given condition.

286 Mathematics—XII ∴ λ = –1/5 is impossible. Case II λ = –1 (3) ⇒ x + y + z – 1 + (–1)(2x + 3y + 4z – 5) = 0 ⇒ –x – 2y – 3z + 4 = 0 ⇒ x + 2y + 3z = 4 ⇒ x + y + z = 1. This is the required plane. 4 2 4/3 Example 35. Find the cartesian and vector equations of the planes passing through the Solution. intersection of the planes (2iˆ + 6ˆj) ⋅ →r + 12 = 0 and (3iˆ − ˆj + 4kˆ) ⋅ →r = 0 which are at unit distance from the origin. Given planes are (2iˆ + 6 ˆj) ⋅ →r + 12 = 0 i.e., (iˆ + 3 ˆj) ⋅ →r + 6 = 0 ...(1) and (3iˆ − ˆj + 4kˆ) ⋅ →r = 0. ...(2) ...(3) (1) ⇒ x + 3y + 6 = 0 ...(4) and (2) ⇒ 3x – y + 4z = 0 ...(5) Let the equation of the required plane be x + 3y + 6 + λ(3x – y + 4z) = 0. ⇒ (1 + 3λ)x + (3 – λ)y + 4λz + 6 = 0 This plane is at a unit distance from the origin. ∴ (1 − 3λ) ⋅ 0 + (3 − λ) ⋅ 0 + (4λ) ⋅ 0 + 6 = 1 (1 + 3λ)2 + (3 − λ)2 + (4λ)2 ⇒ ⎛ 1 + 9λ2 6 − 6λ + 16λ2 ⎞2 =1 ⎜ + 6λ + 9 + λ2 ⎟ ⎜ ⎟ ⎝ ⎠ ⇒ 36 = 26λ2 + 10 ⇒ λ2 = 1 ⇒ λ = ± 1 Case I. λ = 1 ∴ (5) ⇒ x + 3y + 6 + 1⋅(3x – y + 4z) = 0 or 4x + 2y + 4z + 6 = 0 or 2x + y + 2z + 3 = 0 Case II. λ = –1 ∴ (5) ⇒ x + 3y + 6 + (–1)⋅(3x – y + 4z) = 0 or –2x + 4y – 4z + 6 = 0 or x – 2y + 2z – 3 = 0. ∴ The required planes are 2x + y + 2z + 3 = 0 and x – 2y + 2z – 3 = 0. In vector form, these planes take the form (2iˆ + jˆ + 2kˆ ) . → +3=0 and (iˆ − 2 jˆ + 2kˆ ) . → – 3 = 0. r r

Three Dimensional Geometry 287 PRACTICE EXERCISE 5.2 Very Short/Short Answer Type Questions 1. (i) Find the vector equation of the plane which is at a distance of 6 units from the 29 origin and its normal vector from the origin and directed towards the plane is 2iˆ – 3ˆj + 4kˆ . Also find its cartesian form. (NCERT) (ii) Find the vector equation of the plane which is at a distance 5 units from the origin and normal to the vector 2iˆ − 3ˆj + 6kˆ , directed toward the plane. (CBSE 2016) 2. Find the cartesian form of equation of the plane whose vector equation is: (i) (2iˆ − 3 ˆj + 4kˆ) ⋅ →r = 1 (NCERT) (ii) [(s − 2t)iˆ + (3 − t)ˆj + (2s + t)kˆ] ⋅ →r = 15 3. Find the vector form of equation of the plane whose cartesian equation is: (i) x + 2y + 3z = 19 (ii) 5x – y + 3z + 7 = 0. 4. Find the equation of the plane whose intercepts on the coordinate axes are: (i) 2, 3, 4 (NCERT) (ii) 3, –4, 2. (A.I. CBSE 2016) 5. Reduce the following equation to the intercept form and find its intercepts on the coordinates axes: (i) 5x – 3y + 2z = 10 (ii) 2x + y – z = 5. (NCERT) 6. Find the equation of the plane making equal intercepts on the axes and passing through the point (2, 8, 3). 7. Find the value of λ such that the line x − 2 = y − 1 = z+3 is perpendicular to the 9 λ −6 plane 3x – y – 2z = 7. 8. (i) Find the equation of the line passing through the point (1, 2, 3) and perpendicular to the plane (iˆ + 2 ˆj − 5kˆ) ⋅ →r + 9 = 0. (NCERT; A.I. CBSE 2015 C) (ii) Find the equation of the line passing through the point (3, 4, 1) and perpendicular to the plane 3x – y – 5 = 0. 9. (i) Find the equation of the line passing through the point (3, 0, 1) and parallel to the planes x + 2y = 0 and 3y – z = 0. [NCERT (EP)] (ii) Find the vector equation of the line passing through (1, 2, 3) and parallel to the planes (iˆ − ˆj + 2kˆ) ⋅ → =5 and (3iˆ + ˆj + kˆ) ⋅ →r = 6. r (NCERT; A.I. CBSE 2013, 2018 C) 10. In the following cases, find the coordinates of the foot of perpendicular drawn from the origin: (i) x + y + z = 1 (ii) 5y + 8 = 0. (NCERT) 11. (i) Find the coordinates of the foot of perpendicular from the point (2, 3, 7) to the plane 3x – y – z = 7. Also, find the length of the perpendicular. (ii) From the point P(1, 2, 4), a perpendicular is drawn on the plane 2x + y – 2z + 3 = 0. Find the equation, the length and the coordinates of the foot of the perpendicular.

288 Mathematics—XII 12. (i) Find the coordinates of the foot of the perpendicular and the perpendicular distance of the point P(3, 2, 1) from the plane 2x – y + z + 1 = 0. Find also, the image of the point P in the plane. (A.I. CBSE 2012 C) (ii) Find the position vector of the foot of perpendicular and the parpendicular distance from the point P with position vector 2iˆ + 3 ˆj + 4kˆ to the plane (2iˆ + ˆj + 3kˆ) . → = 26. Also, find the image of P in the plane. r (A.I. CBSE 2016) 13. (i) Find the image of the point having position vector iˆ + 3ˆj + 4kˆ in the plane (2iˆ − ˆj + kˆ) ⋅ →r + 3 = 0. [NCERT (EP); A.I. CBSE 2014 C] (ii) Find the image of the point (7, 1, 4) in the plane 2x + 4y – z = 2. 14. (i) Find the image of the point (1, 2, 3) in the plane x + 2y + 4z = 38. (ii) Show that the image of the point (3, –2, 1) in the plane 3x – y + 4z = 2 lies on the plane x + y + z + 4 = 0. 15. (i) Find the distance of the point (–1, –5, –10) from the point of intersection of the line →r = 2iˆ − ˆj + 2kˆ + λ(3iˆ + 4 ˆj + 2kˆ) and the plane (iˆ − ˆj + kˆ) ⋅ →r = 5 . [NCERT (EP); A .I. CBSE 2014 C; CBSE 2018] (ii) Find the distance of the point (2, 12, 5) from the point of intersection of the line → = 2iˆ – 4 ˆj + 2kˆ + λ(3iˆ + 4 ˆj + 2kˆ) and the plane (iˆ − 2 ˆj + kˆ) ⋅ →r = 0. (A.I. CBSE 2014) r 16. (i) Find the distance of the point P(3, 4, 4) from the point, where the line joining the points A(3, –4, –5) and B(2, –3, 1) intersects the plane 2x + y + z = 7. (A.I. CBSE 2015) (ii) Find the coordinates of the point where the line through (3, –4, –5) and (2, –3, 1) crosses the plane 2x + y + z = 7. (NCERT; A.I. CBSE 2012) 17. (i) Find the distance of the point (2, 3, 4) from the plane 3x + 2y + 2z + 5 = 0 measured parallel to the line x + 3 = y − 2 = z . 3 6 2 (ii) Find the distance of the point (1, –2, 3) from the plane x – y + z = 5 measured parallel to the line x – 1 = y – 3 = z+2 . (CBSE 2013 C) 2 3 –6 18. (i) Find the coordinate of point P where the line through A(3, –4, –5) and B(2, –3, 1) crosses the plane passing through three points L(2, 2, 1), M(3, 0, 1) and N(4, –1, 0). Also, find the ratio in which P divides the line segment AB. (CBSE 2016) (ii) Find the coordinates of the point where the line through the points (3, –4, –5), and (2, –3, 1), crosses the plane determined by the points (1, 2, 0), (4, 2, –3) and (0, 4, 3). Also, find the ratio in which this point divides the line joining (3, –4, –5) and (2, –3, 1). (A.I. CBSE 2017) 19. In the following cases, find the distance of each of the given points from the corresponding given plane: Point Plane (i) (0, 0, 0) 2x – 3y + 6z + 21 = 0 (NCERT; A.I. CBSE 2013) (ii) (–6, 0, 0) 2x – 3y + 6z – 2 = 0 (NCERT) (iii) (3, –2, 1) 2x – y + 2z + 3 = 0 (NCERT) (iv) (2, 1, –1) (iˆ − 2ˆj + 4kˆ) . →r = 9 [(NCERT (EP)]

Three Dimensional Geometry 289 (v) (3, 3, 3) (5iˆ + 2 ˆj − 7kˆ) . →r + 9 = 0 (NCERT ; A.I. CBSE 2015) (vi) (2, 5, –3) (6 iˆ − 3 ˆj + 2kˆ) ⋅ →r = 4. Long Answer-I Type Questions 1. (i) Find the vector and cartesian equations of the plane that passes through the point (1, 0, –2) and the normal to the plane is iˆ + ˆj − kˆ . (NCERT) (ii) Find the vector and cartesian equations of the plane that passes through the point (1, 4, 6) and the normal vector to the plane is iˆ − 2ˆj + kˆ . (NCERT) 2. (i) Find the vector and cartesian equations of the plane passing through the point (5, 2, –4) and perpendicular to the line with direction ratios 2, 3, –1. (NCERT) (ii) Find the vector equation of the plane passing through the point (2, 0, –1) and normal to the line joining the points (1, 2, 3) and (3, –1, 6). 3. (i) The foot of the perpendicular drawn from the origin to a plane is (12, –4, –3). Find the equation of the plane. (ii) If O be the origin and the coordinates of P be (1, 2, –3), then find the equation of the plane passing through P and perpendicular to OP. (NCERT) 4. (i) If the line drawn from (4, –1, 2) meets a plane at right angle at the point (–10, 5, 4), then find the equation of the plane. (ii) If the line drawn from the point (–2, –1, –3) meets a plane at right angle at the point (1, –3, 3), find the equation of the plane. [NCERT (EP)] 5. O is the origin and A is (a, b, c). Find the direction cosines of the line OA and the equation of the plane through A and at right angle to OA. [NCERT (EP)] 6. (i) Find the equation of the plane that bisects the line joining the points (2, 3, 4) and (4, 5, 8) and is at right angle to this line. [NCERT (EP)] (ii) Find the equation of the plane that bisects the line joining the points with position vectors ˆj + 2kˆ and 4iˆ + 3ˆj and is at right angle to this line. 7. In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin: (i) z = 2 (NCERT) (ii) x + y + z = 1 (NCERT) (iii) 2x + 3y – z = 5 (NCERT) (iv) 5y + 8 = 0 (NCERT) (v) (iˆ + ˆj + kˆ) ⋅ →r + 4 = 0 (vi) (6iˆ − 3 ˆj − 2kˆ) ⋅ →r + 1 = 0. 8. Find the equation of the plane that passes through the following three points: (i) (2, 1, 0), (3, –2, –2), (3, 1, 7) (ii) (1, 1, –1), (6, 4, –5), (–4, –2, 3) (iii) (1, 1, 0), (1, 2, 1), (–2, 2, –1) (iv) (2, 5, –3), (–2, –3, 5), (5, 3, –3). 9. Find x such that the points A(4, 4, 4), B(5, x, 8), C(5, 4, 1) and D(7, 7, 2) are coplanar. (CBSE 2018 C) 10. A plane meets the coordinates axes at A, B, C such that the centroid of the triangle ABC is (1, –2, 3). Show that the equation of the plane is 6x – 3y + 2z = 18. (CBSE 2014 SP) 11. A variable plane which remains at a constant distance 3p from the origin cuts the coordinate axes at A, B, C. Show that the locus of the centroid of ΔABC is x–2 + y–2 + z–2 = p–2. (A.I. CBSE 2017) 12. Show that the lines x+3 = y − 1 = z − 5 and x+1 = y − 2 = z − 5 are coplanar. Also, −3 1 5 −1 2 5 find the equation of the plane containing these lines. (A.I. CBSE 2013 C)

290 Mathematics—XII 13. If the lines x–1 y–2 z–3 and x–1 y–2 z–3 are perpendicular, find the –3 = –2k = 2 k=1=5 value of k and hence find the equation of the plane containing these lines. (A.I. CBSE 2012) 14. Find the vector and cartesian equations of a plane containing the two lines: →r = 2iˆ + ˆj – 3kˆ + λ (iˆ + 2ˆj + 5kˆ) and →r = 3iˆ + 3 ˆj + 2kˆ + μ(3iˆ – 2 ˆj + 5kˆ). Also show that the line → = 2iˆ + 5 ˆj + 2kˆ + p (3iˆ – 2 ˆj + 5kˆ) lies in the plane. (CBSE 2012 SP) r Find the equation of the plane passing through the given points and parallel to the given line (Q. No. 15–16): 15. (–1, 2, 0), (2, 2, –1); x − 1 = 2y + 1 = z+1 (A.I. CBSE 2015) 1 2 −1 16. (1, 5, 3), (2, 1, 1); x + 1 = 7 − y = z − 1 . 5 1 2 Find the equation of the plane passing through the points with given position vectors and parallel to the given line (Q. No. 17–18): 17. iˆ + ˆj + 3kˆ, 3iˆ + kˆ ; →r = 2iˆ − ˆj + kˆ + λ(iˆ − ˆj + kˆ) 18. 3iˆ – 5kˆ, − iˆ – ˆj − kˆ ; →r = iˆ − ˆj + λ(5iˆ + ˆj + kˆ) . Find the equation of the plane containing the given line and passing through the given point (Q. No. 19–20): 19. x − 1 = y − 2 = z+3 , (2, 3, –2) 20. x = y − 1 = 1 − z , (–1, 0, 2). 2 1 −1 −2 3 1 21. Find the equation of the plane passing through the point P(1, 1, 1) and containing the line →r = (−3 iˆ + ˆj + 5kˆ) + λ(3 iˆ − ˆj − 5kˆ) . Also, show that this plane contains the line →r = (− iˆ + 2 ˆj + 5kˆ) + μ(iˆ − 2ˆj − 5kˆ) . 22. Find the equation of the plane passing through the given point and parallel to the given lines (1, 2, 0), x − 1 = y + 1 = z − 7 , x+2 = y + 1 = z−1. 2 2 2 1 3 2 23. Find the equation of the plane passing through the point with given position vector and parallel to the given lines iˆ + 2 ˆj + 3kˆ, →r = iˆ − ˆj + kˆ + λ(3iˆ − ˆj + 5kˆ), →r = 2iˆ + ˆj − kˆ + μ(iˆ − ˆj + kˆ) 24. Find the equation in the scalar product form of the following planes →r = 2iˆ − ˆj + λ(iˆ + 2 ˆj + 3kˆ) + μ (2iˆ − ˆj − 3kˆ) 25. (i) Find the distance between the parallel planes 2x – y + 2z = 5 and 5x – 2.5y + 5z = 20. (A.I. CBSE 2017) (ii) Find the distance between the planes 2x + 3y + 4z = 4 and 4x + 6y + 8z = 12. (NCERT)

Three Dimensional Geometry 291 26. A plane makes intercepts –6, 3, 4 respectively on the coordinate axes. Find the length of the perpendicular from the origin on it. (CBSE 2014 C) 27. If the points (1, 1, p) and (–3, 0, 1) are equidistant from the plane (3iˆ + 4 ˆj − 12kˆ) ⋅ → + 13 = 0, r then find the value of p. (NCERT) 28. Find the equation of the plane that contains the point (1, –1, 2) and is perpendicular to both the planes 2x + 3y – 2z = 5 and x + 2y – 3z = 8. Hence find the distance of point P(–2, 5, 5) from the plane obtained above. (CBSE 2014) 29. Show that if a plane has intercepts a, b, c and is at a distance of p units from the origin then 1 + 1 + 1 = 1 . (NCERT) a2 b2 c2 p2 30. Find the equation of the plane containing the lines: →r = iˆ + ˆj + λ(iˆ + 2ˆj − kˆ) and →r = iˆ + ˆj + μ(−iˆ + ˆj − 2kˆ) . Find the distance of this plane from origin and also from the points (1, 1, 1) and (2, 1, 4). (A.I. CBSE 2012 C) 31. Find the vector equation of the plane determined by the points A(3, –1, 2), B(5, 2, 4) and C(–1, –1, 6). Also, find the distance of the point P(6, 5, 9) from this plane. (CBSE 2012, 2012 SP, 2013) 32. Find the equation of the plane passing through the points (3, –3, 1) and perpendicular to the line joining the points (3, 4, –1) and (2, –1, 5). Also find the coordinates of the foot of perpendicular and the equation of perpendicular line and the length of perpendicular drawn from origin to the plane. (CBSE 2012 C) 33. Find the distance between the line x − 5 = y − 4 = z −8 and the plane determined by 3 3 1 the points A(2, –2, 1), B(4, 1, 3) and C(–2, –2, 5). (A.I. CBSE 2017 C) Long Answer-II Type Question 1. Find the symmetric form of the lines 2x + 3y + z – 4 = 0 = x + y – 2z – 3 and 5x + 8y – 7z = 0 = 5x – y – z and show that these lines are perpendicular to each other. 2. (i) Find the direction ratios of the normal to the plane passing through the point (2, 1, 3) and the line of intersection of the planes x + 2y + z = 3 and 2x – y – z = 5. (ii) Find the equation of the plane through the intersection of the planes 3x – y + 2z – 4 = 0 and x + y + z – 2 = 0 and the point (2, 2, 1). (NCERT) 3. (i) Find the equation of the plane through the line of intersection of the planes x + y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to the plane x – y + z = 0. Also find the distances of the plane obtained above, from the origin and the point (1, 3, 6). (NCERT ; A.I. CBSE 2014 ; CBSE 2015 C) (ii) Find the equation of the plane which contains the line of intersection of the planes (iˆ + 2 ˆj + 3kˆ) ⋅ → – 4 = 0 and (2iˆ + ˆj − kˆ) ⋅ →r + 5 = 0 and which is perpendicular to the r plane (5iˆ + 3 ˆj − 6kˆ) ⋅ → + 8 = 0. (NCERT; CBSE 2013) r 4. (i) Find the equation of the plane passing through the line of intersection of the planes 2x + y – z = 3 and 5x – 3y + 4z + 9 = 0 and parallel to the line x − 1 = y − 3 = z−5 . 2 4 5 (A.I. CBSE 2011)

292 Mathematics—XII (ii) Find the equation of the plane passing through the intersection of the planes 4x – y + z = 0 and x + y – z = 4 and parallel to the line with direction ratios 2, 1, 1. Find also the perpendicular distance of (1, 1, 1) from this plane. 5. (i) Find the equation of the plane through the line of interesection of the planes 3x – 4y + 5z = 10, 2x + 2y – 3z = 4 and parallel to the line x = 2y = 3z. (ii) Find the equation of the plane passing through the intersection of the planes (iˆ + ˆj + kˆ) ⋅ → = 1 and (2iˆ + 3 ˆj − kˆ) ⋅ → + 4 = 0 and parallel to the x-axis. r r (NCERT; A.I. CBSE 2014 C) 6. (i) Find the cartesian and vector equations of the planes passing through the intersection of the planes (iˆ + 3 ˆj ) ⋅ → – 6 = 0 and (3iˆ – ˆj – 4kˆ) ⋅ → = 0 which are at unit distance r r from the origin. (A.I. CBSE 2013) (ii) Find the cartesian and vector equations of the plane through the line of intersection of the planes (iˆ + 3 ˆj ) ⋅ → + 6 = 0 and (3iˆ – ˆj – 4kˆ ) ⋅ → = 0 which is at a unit distance r r from the origin. [NCERT (EP); A.I. CBSE 2013 C] 7. Find the vector and cartesian equations of the plane passing through the line of ( ) ( )intersection of the planes →r . 2iˆ + 2ˆj − 3kˆ = 7 , →r . 2iˆ + 5ˆj + 3kˆ = 9 such that the intercepts made by the plane on the x-axis and z-axis are equal. (A.I. CBSE 2015 C) Answers Very Short/Short Answer Type Questions (ii) (2iˆ − 3 ˆj + 6kˆ) ⋅ →r = 35 1. (i) (2iˆ − 3 ˆj + 4kˆ) ⋅ →r = 6, 2x – 3y + 4z = 6 (ii) (s – 2t)x + (3 – t)y + (2s + t)z = 15 2. (i) 2x – 3y + 4z = 1 3. (i) (iˆ + 2 ˆj + 3kˆ) ⋅ →r = 19 (ii) (5i − j + 3k) ⋅ →r = –7 4. (i) 6x + 4y + 3z = 12 (ii) 4x – 3y + 6z = 12 5. (i) 2, – 10 ,5 (ii) 5 , 5, − 5 3 2 6. x + y + z = 13 7. –3 8. (i) →r = iˆ + 2 ˆj + 3kˆ + λ(iˆ + 2 ˆj − 5kˆ) (ii) x − 3 = y−4 = z − 1 3 –1 0 9. (i) x−3 = y = z − 1 (ii) → = iˆ + 2 ˆj + 3kˆ + λ(–3iˆ + 5 ˆj + 4kˆ) −2 1 3 r 10. (i) ⎛ 1 , 1 , 1 ⎞ (ii) ⎛ 0, − 8 , 0 ⎞ ⎜⎝ 3 3 3 ⎟⎠ ⎝⎜ 5 ⎠⎟ 11. (i) (5, 2, 6), 11 units (ii) x−1 = y−2 = z−4 , 1 unit, ⎛ 11 , 19 , 34 ⎞ 2 1 −2 3 ⎜⎝ 9 9 9 ⎟⎠ 12. (i) (1, 3, 0), 6 units, (–1, 4, –1) (ii) (3, 7/2, 11/2), 14/2 units, (4, 4, 7)

Three Dimensional Geometry 293 13. (i) (–3, 5, 2) (ii) ⎛ 33 , − 25 , 36 ⎞ ⎝⎜ 7 7 7 ⎟⎠ 14. (i) (3, 6, 11). 16. (i) 7 units (ii) (1, –2, 7) 15. (i) 13 units (ii) 13 units 18. (i) (1, –2, 7), 2 : 1 external 17. (i) 7 units (ii) 1 unit 19. (i) 3 units (ii) (15/4, –19/4, –19/2); 3 : 7 external. (ii) 2 units (iii) 13 units 3 (iv) 13 21 units (v) 9 78 units (vi) 13 units 21 78 7 Long Answer-I Type Questions 1. (i) (iˆ + ˆj − kˆ) ⋅ → = 3, x+y–z=3 (ii) (iˆ − 2 ˆj + kˆ) ⋅ → = –1, x – 2y + z = –1 r r 2. (i) (2iˆ + 3 ˆj − kˆ) ⋅ → = 20, 2x + 3y − z − 20 = 0 r (ii) (2iˆ – 3 ˆj + 3kˆ) ⋅ → =1 r 3. (i) 12x – 4y – 3z = 169 (ii) x + 2y – 3z = 14 (ii) 3x – 2y + 6z = 27 4. (i) 7x – 3y – z + 89 = 0 c ; ax + by + cz = a2 + b2 + c2 5. a , b , a2 + b2 + c2 a2 + b2 + c2 a2 + b2 + c2 6. (i) x + y + 2z = 19 (ii) (2iˆ + ˆj − kˆ) ⋅ →r = 5 7. (i) 0, 0, 1; 2 (ii) 1 , 1 , 1 ; 1 3 3 3 3 (iii) 2 , 3 , − 1 ; 5 (iv) 0, 1, 0; 8 14 14 14 14 5 (v) 1 , 1 , 1 ; 4 (vi) − 6 , 3 , 2 ; 1 3 3 3 3 7 7 7 7 8. (i) 7x + 3y – z = 17 (ii) Given points are collinear, so there will be infinitely many planes passing through given points. (iii) 2x + 3y – 3z = 5 (iv) 2x + 3y + 4z = 7 9. 7. 12. x – 2y + z = 0 13. k = 2, 22x – 19y – 5z + 31 = 0 14. (10 iˆ + 5 ˆj – 4kˆ) ⋅ →r = 37, 10x + 5y – 4z = 37. 15. x + 2y + 3z – 3 = 0 16. 10x + 12y – 19z = 13 17. (3iˆ + 4 ˆj + kˆ) ⋅ → = 10 r 18. (5iˆ − 24 ˆj − kˆ) ⋅ →r = 20 19. 2x – 3y + z + 7 = 0 20. 2x + 3y + 5z = 8 21. x – 2y + z = 0. 22. x + y – 2z = 3 23. (2iˆ + ˆj − kˆ) ⋅ →r = 1 24. (3iˆ − 9 ˆj + 5kˆ) ⋅ → = 15 25. (i) 1 unit (ii) 2 29 unit 27. 1 or 7/3 29 r 26. 12 29 units 29

294 Mathematics—XII 28. 5x – 4y – z – 7 = 0, 42 units 30. x – y – z = 0, 0 unit, 3/3 unit, 3 units 31. (3iˆ – 4 ˆj + 3kˆ) ⋅ →r = 19, 3 34/17 units 32. x + 5y – 6z + 18 = 0, ⎛ – 9 , – 45 , 54 ⎞ , x = y = z , 9 2 units ⎝⎜ 31 31 31 ⎠⎟ 1 5 –6 31 33. 3 34 units. 17 Long Answer-II Type Question 1. x − 5 = y+2 = z − 0 , x − 0 = y − 0 − z − 0 . 7 −5 1 1 2 3 2. (i) 13, 6, 1 (ii) 7x – 5y + 4z = 8 3. (i) x – z + 2 = 0, 2 units, 3 2/2 units (ii) (33iˆ + 45 ˆj + 50kˆ) ⋅ →r = 41 4. (i) 7x + 9y – 10z – 27 = 0 (ii) 5y – 5z – 16 = 0, 82 units 5 5. (i) x – 20y + 27z = 14 (ii) ( ˆj − 3kˆ) ⋅ → +6=0 r 6. (i) 2x + y – 2z – 3 = 0, x – 2y – 2z + 3 = 0, (2iˆ + ˆj − 2kˆ) ⋅ → – 3 = 0, r (iˆ – 2 ˆj – 2kˆ) ⋅ →r + 3 = 0 (ii) 2x + y – 2z + 3 = 0, x – 2y – 2z – 3 = 0, (2iˆ + ˆj − 2kˆ) ⋅ →r + 3 = 0, (iˆ – 2 ˆj – 2kˆ) ⋅ →r – 3 = 0 7. (12iˆ + 27 ˆj + 12kˆ) ⋅ → = 52, 12x + 27 y + 12z = 52 . r HINTS/SOLUTIONS Very Short/Short Answer Type Questions 12. (i) If Q and R be the foot of the perpendicular from P and the image of P respectively, then Q is the midpoint of PR. Long Answer I Type Questions 11. Let the equation in normal form be lx + my + nz = 3p. ∴ A = (3p/l, 0, 0), B = (0, 3p/m, 0), C = (0, 0, 3p/n) If (α, β, γ) is the centroid of ΔABC, then α = p/l, β = p/m, γ = p/n. Now use l2 + m2 + n2 = 1. 27. Given plane is 3x + 4y – 12z + 13 = 0. We have 3(1) + 4(1) − 12p + 13 = 3(− 3) + 4(0) − 12(1) + 13 . ⇒ 9 + 16 + 144 9 + 16 + 144 |20 – 12 p| = 8 ⇒ 20 – 12p = ± 8.

Three Dimensional Geometry 295 CASE STUDY/SOURCE BASED INTEGRATED MCQs Example 1. A student design a cube with each side Z of length a units and name it OAREFCDG, taking O as origin. Based on the above information, answer the following GF questions: DC (i) The coordinates of the vertex F are: (a) (0, a, a) (b) (a, 0, 0) (c) (a, a, 0) (d) (a, a, a) O E Y A B (ii) Length of diagonal OC is (a) a units (b) 3a units (c) 3 a units (d) 2 a units X (iii) Direction cosines of the line AF are (a) 1 , 1 , 1 (b) − 1, 1, 1 333 3 3 3 (c) 1, 1 , − 1 (d) 1 , − 1, 1 3 3 3 3 3 3 (iv) Direction ratios of GB are (a) 1, –1, 1 (b) –1, 1, 1 (c) 1, 1 –1 (d) 1, 1, 1 (v) Co-ordinates of the vertex G are: (a) (a, a, a) (b) (a, – a, a) (c) (0, a, 0) (d) (0, 0, a) Sol. (i) Here point F is (0, a, a) Hence the correct option is (a). (ii) Points of O(0, 0, 0) and C (a, a, a) ∴ OC = (a − 0)2 + (a − 0)2 + (a − 0)2 = a2 + a2 + a2 = 3a2 = 3 a units A(a, 0, 0) F(0, a, a) Hence the correct option is (c). (iii) As co-ordinates of A (a, 0, 0) ∴ Direction cosines of the line AF 0−a , a−0 , a−0 AF AF AF ∴ AF = (0 − a)2 + (a − 0)2 + (a − 0)2 = a2 + a2 + a2 = 3 a −a , a a or − 1 , 1 , 1 3a 3a , 3a 3 3 3 Hence the correct option is (b). (iv) Now a – 0, a – 0, 0 – a a, a, – a or 1, 1, – 1 Hence the correct option is (c). G(0, 0, a) B(a, a, 0) (v) Clearly, the correct option is (d). 295

296 Mathematics—XII Example 2. ILf 1a1anb1dcL12anredspae2,ctbi2v,ecl2ya. re direction ratios of L1 two lines say Then, L1 || L2 if a1 = b1 = c1 L2 a2 b2 c2 and L1 ⊥ L2 if a1a2 + b1b2 + c1c2 = 0 L1 L2 Based on the above information, answer the following questions: (i) If l1, m1 , n1 and l2, m2, n2 are direction cosines of L1 and L2 respectively, then L1 will be parallel to L2 iff (a) l1 = m1 = n1 (b) m1n1 + m2n2 + l1l2 = 0 l2 m2 n2 (c) l1l2 + m1m2 + n1n2 = 0 (d) l1m2 + m1l2 + n1n2 = 0 (ii) wIf ill1l mbe1 pn1erapnedndl2i,cmul2a, rnt2oaLre2, the direction cosines of L1 and L2 respectively, then L1 iff (a) l1m2 + m1l2 + n1n2 = 0 (b) l1l2 + m1m2 + n1n2 = 0 (c) l1 = m1 = n1 (d) l1m1 + m2n1 + l1n2 = 0 l2 m2 n2 (iii) The coordinates of the foot of the perpendicular drawn from the point A (1, 2, 1) to the line joining B (1, 4, 6) and C (5, 4, 4) are (a) (4, 3, 5) (b) (3, 4, 5) (c) (2, 4, 5) (d) (1, 2, 1) (iv) The direction ratios of the line which is perpendicular to the lines with direction ratios proportional to (4, –3, 5) and (3, 4, 5). (a) –35, –5, 25 (b) 35, 5, –25 (c) –35, –5, –25 (d) –35, 5, 25 (v) The line x − 5 = 2 − y = 1−z and x = y + 1 = 1−z are 7 5 −1 1 2 −3 (a) Parallel (b) Perpendicular (c) Skew lines (d) non-intersecting Sol. (i) Since direction ratios are proportional to direction cosines therefore L1 will be parallel l1 m1 n1 to L2 iff l2 = m2 = n2 . Hence the correct option is (a). (ii) Since direction ratios are proportional to direction cosine therefore L1 will be perpendicular to L2 if l1l2 + m1m2 + n1n2 = 0 Hence the correct option is (b). (iii) Let M be the root of the perpendicular drawn from the point A (1, 2, 1) Given co-ordinate of point B and C are B (1, 4, 6) and C (5, 4, 4) Equation of line joining B and C is x−1 = y−4 = z−6 5−1 4−4 4−6

Three Dimensional Geometry 297 x−1 = y−4 = z−6 = λ (say) ...(i) 4 0 −2 x = 4λ + 1, y = 4, z = –2λ + 6 Co-ordinate of M are (4λ + 1, 4, –2λ + 6) The direction ratio of AM are 4λ + 1 – 1, 4 – 2, –2λ + 6 – 1 i.e., 4λ, 2, –2λ + 5 The direction ratio of line (i) are 4, 0, –2 Since AM is perpendicular to the given line ∴ 4 (4λ) + (0) (2) + (–2) (–2λ + 5) = 0 ⇒ 16λ + 4λ – 10 = 0 ⇒ 20λ = 10 ⇒ λ= 10 = 1 Hence co-ordinate of M are 20 2 ⎛ 4 × 1 + 1, 4, − 2 × 1 + 6⎞⎠⎟ ⎝⎜ 2 2 (3, 4, 5) Hence, the correct option is (b). (iv) Let →b 1 = 4iˆ − 3ˆj + 5kˆ and → = 3iˆ + 4ˆj + 5kˆ b2 → →→ iˆ ˆj kˆ b = b1 × b2 = 4 −3 5 3 45 = iˆ (–15 – 20) – ˆj (20 – 15) + kˆ (16 + 9) = –35 iˆ – 5 ˆj + 25 kˆ ∴ Direction ratios of line are –35, –5, 25 Hence the correct option is (a). (v) Given lines x− 5 = 2−y = 1−z and x = y+1 = 1−z 7 5 −1 7 2 −3 x− 5 = y−2 = z−1 ...(i) 7 −5 1 x − 0 = y + 1 = z − 1 ...(ii) 1 2 3 Direction ratios of line (i) and (ii) are 2, –5, 1 and 1, 2, 3 respectively then, (7) (1) + (–5) (2) + (1) (3) = 7 – 10 + 3 = 0 ∴ Given lines are perpendicular to each other. Hence the correct option is (b). Example 3. A pole stands at the top of a hill. Consider the surface on which pole stand as a plane having points P (0, 1, 2), Q (3, 4, –1) and R (2, 4, 2) on it. The pole is tied with 3 cables from the point P, Q and R such that it stand vertically on the ground. The peak of the pole is at the point (6, 5, 9), as shown in the figure. Based on the above information, answer the following questions:

298 Mathematics—XII (6, 5, 9) P(0, 1, 2) Q(3, 4, –1) R(2, 4, 2) (i) The equation of the plane passing through the points P, Q and R is (a) 4x – 3y + 3z = 0 (b) 3x – 2y + z = 0 (c) 3x – 4y + z = 0 (d) 4x – 3y + z = 0 (ii) The height of the pole from the ground is (a) 5 units (b) 5 units (c) 12 units (d) 17 units 14 14 (iii) The equation of line of perpendicular drawn from the peak of pole to the ground is (a) x−6 = y−6 = z−9 (b) x − 6 = y − 6 = z − 9 3 2 2 3 2 1 (c) x − 6 = y−5 = z − 9 (d) x − 6 = y − 5 = z − 9 3 −2 1 1 2 2 (iv) The co-ordinates of foot of perpendicular drawn from the peak of pole to the ground are (a) ⎛ 33 , 109 , 104 ⎞ (b) ⎛ 33 , 104 , 109 ⎞ ⎜⎝ 14 14 14 ⎟⎠ ⎝⎜ 14 14 14 ⎠⎟ (c) ⎛ 33 , −104 , 109 ⎞ (d) ⎛ −33 , 104 , −105⎞ ⎜⎝ 14 14 14 ⎠⎟ ⎜⎝ 14 14 14 ⎠⎟ (v) The area of ΔABC is (a) 1 14 sq. units (b) 14 sq. units 2 (c) 2 14 sq. units (d) 3 14 sq. units Sol. (i) P (0, 1, 2), Q (3, 4, –1), R (2, 4, 2) 2 The equation of a plane passing through three P, Q, R non-collinear points. x − x1 y − y1 z − z1 =0 x2 − x1 y2 − y1 z2 − z1 x3 − x1 y3 − y1 z3 − z1 x−0 y−1 z−2 x y−1 z−2 3 − 0 4 − 1 −1 − 2 = 0 ⇒ 2−0 4−1 2−2 3 3 −3 = 0 23 0

Three Dimensional Geometry 299 x (0 + 9) – (y – 1) (0 + 6) + (z – 2) (9 – 6) = 0 ⇒ 9x – 6y + 6 + 3z – 6 = 0 ⇒ 9x – 6y + 3z – 6 + 6 = 0 ⇒ 3x – 2y + z = 0 Hence the correct option is (b). (ii) The height of pole = Perpendicular distance from the point (6, 5, 9) to the plane 3x – 2y + z = 0 = 3×6−2×5+9 = 18 − 10 + 9 = 17 units (3)2 + (−2)2 + (1)2 9+4+1 14 Hence the correct option is (d). (iii) Direction ratios of perpendicular are < 3, –2, 1 > Θ Perpendicular is parallel to the normal to the plane Since, perpendicular is passing through the point (6, 5, 9) Therefore its equation is x−6 = y−5 = z−9 3 −2 1 Hence the correct option is (c). (iv) We have x−6 = y−5 = z−9 = λ (say) 3 −2 1 ⇒ x = 3λ + 6, y = –2λ + 5, z = λ + 9 Co-ordinates of foot of perpendicular are (3λ + 6, −2λ + 5, λ + 9) Since, this point lies on the plane 3x – 2y + z = 0 ⇒ 3 (3λ + 6) – 2 (–2λ + 5) + (λ + 9) = 0 ⇒ 9λ + 18 + 4λ − 10 + λ + 9 = 0 ⇒ 14λ + 17 = 0 ⇒ λ= − 17 14 Now, x= 3× −17 +6 = −51 + 84 = 33 14 14 14 y= − 2 × − 17 +5 = 34 + 70 = 104 14 14 14 z= λ+9= −17 + 9 = −17 + 126 = 109 14 14 14 Co-ordinates of foot of perpendicular are ⎛⎜⎝ 33 , 104 , 109 ⎟⎞⎠ 14 14 14 Hence the correct option is (b). (v) P (0, 1, 2), Q (3, 4, –1), R (2, 4, 2) → = Position vector of Q – Position vector of P PQ = (3iˆ + 4ˆj − kˆ) – (0iˆ + ˆj + 2kˆ) = (3iˆ + 3ˆj − 3kˆ) → = Position vector of R – Position vector of P PR = (2iˆ + 4ˆj + 2kˆ) – (0iˆ + ˆj − 2kˆ) = (2iˆ + 3ˆj + 0kˆ)

300 Mathematics—XII Area of ΔPQR = 1 →→ = 1 iˆ ˆj kˆ 2 PQ × PR 2 3 3 −3 2 3 0 = 21|iˆ(0 + 9) − ˆj(0 + 6) + kˆ(9 − 6)| = 21|9iˆ − 6 ˆj + 3kˆ| = 1 (9)2 + (−6)2 + (3)2 = 1 81 + 36 + 9 = 126 2 2 2 = 3 14 sq. units 2 Hence the correct option is (d). Example 4. Two cars A and B are running at the speed more than allowed speed on the road along the lines → = λ(3iˆ + 2 ˆj + 6kˆ) and → = (7iˆ − 6kˆ) + μ (iˆ + 2 ˆj + 2kˆ) , respectively r r Based on the above information, answer the following questions: (i) The cartesian equation, of the line along which car A is running, is (a) x = y = z (b) x = y = z 3 2 6 −3 2 6 (c) x − 3 = y = z − 1 (d) None of these. 2 2 6 (ii) The cartesian equation of the line along which car B is running is (a) x = y − 6 = z + 1 (b) x − 1 = y − 0 = z − 6 7 2 2 7 2 2 (c) x − 7 = y = z + 6 (d) x − 7 = y − 2 = z + 6 1 2 2 2 2 2 (iii) The direction cosines of the line along which car A is running, are (a) < 1, 2, –1 > (b) 3 , 2 , 6 7 7 7 (c) 1 , 2 , −1 (d) 1 , −2 , 1 6 6 6 6 6 6 (iv) The direction ratios of the line along which car B is running, are (a) < 1, 2, 2 > (b) < 2, 1, 1 > (c) < 2, 1, 0 > (d) < 1, 0, 2 > (v) The shortest distance between the given lines are (a) 3 5 units (b) 5 units (c) 2 5 units (d) 4 5 units Sol. (i) The line along which car A is running → = λ(3iˆ + 2 ˆj + 6kˆ) r Let →r = xiˆ + yˆj + zkˆ (xiˆ + yˆj + zkˆ) = λ(3iˆ + 2 ˆj + 6kˆ) x = 3λ, y = 2λ, z = 6λ ∴ x = λ, y = λ, z =λ or x = y = z 3 2 6 3 2 6 Hence the correct option is (a).

Three Dimensional Geometry 301 (ii) The line along which car B is running →r = (7iˆ − 6kˆ) + μ (iˆ + 2 ˆj + 2kˆ) Let → = xiˆ + yˆj + zkˆ r ∴ xiˆ + yˆj + zkˆ = (7iˆ − 6kˆ) + μ (iˆ + 2ˆj + 2kˆ) = (7 + μ)iˆ + 2μˆj + (−6 + 2μ)kˆ On equating the coefficients of iˆ, ˆj and kˆ , then x = 7 + μ, y = 2μ z = –6 + 2μ x – 7 = μ, y =μ, z+6 =μ 2 2 On eleminating μ, we obtain x−7 = y = z+6 1 2 2 Hence the correct option is (c). (iii) Clearly, direction ratios of the required line are < 3, 2, 6 >. ∴ Direction cosines are < 3, 2,6 > (3)2 + (2)2 + (6)2 (3)2 + (2)2 + (6)2 (3)2 + (2)2 + (6)2 < 9 + 3 + 36 , 9 + 2 + 36 , 6> or < 3 , 2 , 6 > 4 4 9 + 4 + 36 7 7 7 Hence the correct option is (b). (iv) Direction ratios of the line along which car B are < 1, 2, 2 > Hence the correct option is (a). (v) Line I Line II → = λ (3iˆ + 2 ˆj + 6kˆ) → = (7iˆ − 6kˆ) + μ (iˆ + 2 ˆj +2kˆ) r r Here a1 = 0iˆ + 0 ˆj + 0kˆ , → = 3iˆ + 2 ˆj + 6kˆ b1 a2 = 7iˆ − 6kˆ , → = iˆ + 2 ˆj + 2kˆ b2 → → = 7iˆ − 6kˆ a2 − a1 →→ iˆ ˆj kˆ = iˆ (4 − 12) − ˆj (6 − 6) + kˆ (6 − 2) b1 × b2 = 3 2 6 122 = − 8iˆ − 0 ˆj + 4kˆ →→ = (−8)2 + (0)2 + (4)2 = 64 + 0 + 16 = 80 = 4 5 units b 1 × b2 Hence the correct option is (d).

302 Mathematics—XII PRACTICE EXERCISE 5.3 1. Let A (2, 3, 5), B (–1, 2, 4) and C (5, 4, 6) be three points in space. Based on the above information, answer the following questions: (i) Direction ratios of the line AB are (a) 1 : 5 : 9 (b) 3 : 1 : 1 (c) 3 : 0 : 1 (d) None of these (ii) Direction ratios of the line BC are (a) 6 : 2 : 2 (b) 3 : 1 : 1 (c) 3 : 1 : 1 (d) All are true 3 3 3 (iii) The direction cosines of the line AC are: (a) 3, 1, 1 (b) 3 11, 11, 11 (c) 3 , 1 , 1 (d) None of these 11 11 11 (iv) Area of triangle ABC is equal to : (a) 4 sq. units (b) 2 sq. units (c) 0 sq. unit (d) None of these 2. Points A (4, 5, 10), B (2, 3, 4) and C (1, 2, –1) are three vertices of a parallelogram ABCD Based on the above information, answer the following questions: (i) Vector equation of the line AB is given by: (a) → = 4iˆ + 5ˆj + 10 ˆj + λ (2iˆ + 3ˆj + 4kˆ) D C(1, 2, –1) r (b) → 4iˆ + 5ˆj + 10ˆj + λ (6iˆ + 8ˆj + 14kˆ) r= (c) →r = 4iˆ + 5ˆj + 10kˆ + λ (iˆ + ˆj + 3kˆ) A(4, 5, 10) B(2, 3, 4) (d) None of these (ii) Cartesian equation the line BC is given by (a) x − 1 = y − 2 = z + 1 (b) x − 2 = y − 3 = z − 4 2 3 4 1 1 5 (c) x − 2 = y − 3 = z − 4 (d) None of these 3 5 3 (iii) Mid-point of AC is given by: (a) (3, 4, 7) (b) ⎛ 5 , 7 , 9⎞ ⎜⎝ 2 2 2 ⎠⎟ (c) (5, 7, 9) (d) None of these (iv) The co-ordinates of the vertex D are given by: (a) (3, –4, 5) (b) (3, 4, –5) (c) (3, 4, 5) (d) None of these

Three Dimensional Geometry 303 3. Points A (4, 7, 8), B (2, 3, 4), C (–1, –2, 1) and D (1, 2, 5) are four points in space. Based on the above information, answer the following questions: (i) Direction cosines of AB are: (a) –2, –4, –4 (b) 1, 2, 2 (c) 1 , 2 , 2 (d) None of these 3 3 3 (b) 1, 2, 2 (ii) Direction cosines of CD are (a) 2, 4, 4 (c) 1 , 2 , 2 (d) None of these 3 3 3 (iii) Lines AB and CD are: (b) perpendicular to each other (d) None of these (a) parallel (c) inclined at 30° (iv) Vector equation of line AB is: (a) → = 4iˆ + 7 ˆj + 8kˆ + λ(2iˆ + 3ˆj + 4kˆ) r (b) → = 2iˆ + 3 ˆj + 4kˆ + λ (4iˆ + 7 ˆj + 8kˆ) r (c) → = 4iˆ + 7 ˆj + 8kˆ + λ (iˆ + 2 ˆj + 2kˆ) r (d) None of these (v) Cartesian equations of line CD are: (a) x − 1 = y − 2 = z + 1 (b) x + 1 = y + 2 = z − 1 1 2 5 1 2 2 (c) x − 1 = y − 2 = z−5 (d) None of these 1 2 −1 4. Consider the line x + 1 = y−3 = z and the point P(1, 2, –3). 2 −2 −1 Based on the above information, answer the following questions: (i) Direction cosines of the given line are: (a) − 1 , 3 , 0 (b) 2, –2, –1 10 10 (c) − 2 , 2 , 1 (d) None of these 3 3 3 (ii) The point (1, –3, 0): (a) lies on the given line (b) does not lie on the line (c) is at a distance 5 units from the line (d) None of these (iii) The direction cosines of the line joining P and Q (0, 0, 0) are: (a) 1, 2, –3 (b) 1, 2 , − 3 14 14 14 (c) 1 , 2 , − 3 (d) None of these 6 6 6

304 Mathematics—XII (iv) The foot of perpendicular from P to the given line is: (a) (1, –1, 1) (b) (–1, 1, 1) (c) (1, 1, –1) (d) None of these (v) The image of P in the given line is: (a) (1, 0, 1) (b) (1, 1, 1) (c) (1, 1, 0) (d) None of these 5. Consider the lines l1 : → = 3iˆ + 2 ˆj − 4kˆ + λ (iˆ + 2 ˆj + 2kˆ) r and l2 : → = 5iˆ − 2ˆj + μ (3iˆ + 2ˆj + 6kˆ) r Based on the above information, answer the following questions: (i) Line l1 passes through: (b) (–3, –2, 4) (a) (1, 2, 2) (c) (3, 2, –4) (d) None of these (ii) The direction cosines of the line l2 are: (a) 3, 2, 6 (b) 3 , 2 , 6 7 7 7 (c) 5 , − 2 ,0 (d) None of these 29 29 (iii) The lines l1 and l2 are: (b) parallel (a) not intersecting (c) perpendicular to each other (d) intersecting at (–1, –6, –12) (iv) The shortest distance between the given lines is: (a) 2 units (b) 0 unit (c) 2 units (d) None of these 17 6. Consider the points A (– 2, 6, –6), B (–3, 10, –9), C (–5, 0, –6) and D (1, 2, 3) in space. Based on the above information, answer the following questions: (i) Equation of the line AB is: (a) → = − 2iˆ + 6ˆj − 6kˆ + λ (− 3iˆ + 10ˆj − 9kˆ) r (b) → = − 3iˆ + 10ˆj − 9kˆ + λ (− 2iˆ + 6ˆj − 6kˆ) r (c) →r = − 2iˆ + 6ˆj − 6kˆ + λ (− iˆ + 4 ˆj − 3kˆ) (d) None of these (ii) Lines AB and CD are: (a) parallel (b) perpendicular to each other (c) inclined at 45° to each other (d) None of these (iii) Equation of the plane passing through A, B and C is given by: (a) 2x + y + 2z = 2 (b) 2x – y + 2z = 2 (c) 2x – y – 2z = 2 (d) None of these

Three Dimensional Geometry 305 (iv) Distance of the point D from the plane through A, B and C is equal to: (a) 8 unit (b) 8 units 9 3 (c) 2 unit (d) None of these 3 (v) Which of the following is true: (a) Point D lies on the plane through A, B and C. (b) Point D does not lie on the plane through A, B and C. (c) Distance of origin through the plane through A, B and C is equal to: (a) 2 unit (b) 4 units 3 3 (c) 8 units (d) None of these 3 7. A cricket match is organised between two clubs A and B for which a team from each club is chosen. Remaining players of club A and club B are respectively sitting on the plane represented by the equation → . → − → → = 3 and →→ + → → = 8, to r (2 i j+ k) r .(i 3j +2 k) cheer the team of their own clubs. Based on the above information, answer the following questions: (i) The Cartesian equation of the plane on which players of club A are seated is (a) 2x – y + z = 3 (b) 2x – y + 2z = 3 (c) 2x – y + z = –3 (d) x – y + z = 3 (ii) The magnitude of the normal to the plane on which players of club B are seated, is (a) 15 (b) 14 (c) 17 (d) 20 (iii) The intercept form of the equation of the plane on which players of club B are seated is (a) x + y + z = 1 (b) x + y + z =1 8 8 2 5 8 3 3 3 (c) x + y + z = 1 (d) x + y + z =1 8 8 4 8 7 2 3

306 Mathematics—XII (iv) Which of the following is a player of club B? (a) Player sitting at (1, 2, 1) (b) Player sitting at (0, 1, 2) (c) Player sitting at (1, 4, 1) (d) Player sitting at (1, 1, 2) (v) The distance of the plane, on which players of club B are seated, from the origin is (a) 8 units (b) 6 units 14 14 (c) 7 units (d) 9 units 14 14 8. The Indian coast guard, while patrolling, saw a suspicious boat with people. They were nowhere looking like fishermen. The coast guard were closely observing the movement of the boat for an opportunity to seize the boat. They observed that the boat is moving along a planar surface. At an instant of time, the coordinates of the position of the coast guard helicopter and the boat is (1, 3, 5) and (2, 5, 3) respectively. Based on the above information, answer the following questions: (i) If the line joining the positions of the helicopter and the boat is perpendicular to the plane in which the boat moves, then the equation of the plane is (a) –x + 2y – 2z = 6 (b) x + 2y + 2z = 6 (c) x + 2y – 2z = 6 (d) x – 2y – 2z = 6 (ii) If the coast guard decide to shoot the boat at that given instant of time, then what is the distance (in meters) that the bullet has to travel? (a) 5 cm (b) 3 cm (c) 6 cm (d) 4 cm (iii) If the coast guard decides to shoot the boat at that given instant of time, when the speed of bullet is 36 m/sec, then what is the time taken for the bullet to travel and hit the boat? (a) 1 seconds (b) 1 seconds 8 14 (c) 1 seconds (d) 1 seconds 10 12 (iv) At that given instant of time, the equation of line passing through the positions of the helicopter and boat is

Three Dimensional Geometry 307 (a) x − 1 = y − 3 = z−5 (b) x − 1 = y + 3 = z−5 1 2 −2 2 1 −2 (c) x+1 = y−3 = z−5 (d) x − 1 = y+3 = z + 5 −2 −1 −2 2 −1 2 (v) At a different instant of time, the boat moves to a different position along the planar surface. What should be the co-ordinates of the location of the boat if the coast guard shoots the bullet along the line whose equation is x = y − 1 = z − 2 for the bullet to hit the boat? 1 2 1 (a) ⎛ −8 , 19 , −14 ⎞ (b) ⎛ 8 , −19 , −14 ⎞ ⎝⎜ 3 3 3 ⎠⎟ ⎝⎜ 3 3 3 ⎠⎟ (c) ⎛ 8 , −19 , 14 ⎞ (d) None of these ⎜⎝ 3 3 3 ⎟⎠ 9. The equation of motion of a missile are x = 3t, y = –4t, z = t, where the time ’t’ is given in seconds, and the distance is measured in kilometres. Based on the above information, answer the following questions: (i) What is the path of the missile? (a) Straight line (b) Parabola (c) Circle (d) Ellipse (ii) Which of the following points lie on the path of the missile? (a) (6, 8, 2) (b) (6, –8, –2) (c) (6, –8, 2) (d) (–6, –8, 2) (iii) At what distance will the rocket be from the starting point (0, 0, 0) in 5 seconds? (a) 550 km (b) 650 km (c) 450 km (d) 750 km (iv) If the position of rocket at a certain instant of time is (5, –8, 10), then what will be the height of the rocket from the ground? (The ground is considered as the xy-plane). (a) 12 km (b) 11 km (c) 20 km (d) 10 km (v) At a certain instant of time, if the missile is above the sea level, where the equation of the surface of sea is given by 2x + y + 3z = 1 and the position of the missile at that instant of time is (1, 1, 2), then the image of the position of the rocket in the sea is (a) ⎛ −9 , −1 , −10 ⎞ (b) ⎛ 9 , −1 , −10 ⎞ ⎜⎝ 7 7 7 ⎟⎠ ⎜⎝ 7 7 7 ⎠⎟ (c) ⎛ −9 , 1 , −10 ⎞ (d) ⎛ −9 , −1 , 10 ⎞ ⎝⎜ 7 7 7 ⎟⎠ ⎜⎝ 7 7 7 ⎟⎠

308 Mathematics—XII 10. Suppose the floor of a hotel is made up of mirror polished Salvatore stone. There is a large crystal chandelier attached to the ceilling of the hotel room. Consider the floor of the hotel room as a plane having the equation x – y + z = 4 and the crystal chandelier is suspended at the point (1, 0, 1) Based on the above information, answer the following questions: (i) Find the direction ratios of the perpendicular from the point (1, 0, 1) to the plane x – y + z = 4. (a) (–1, –1, 1) (b) (1, –1, –1) (c) (–1, –1, –1) (d) (1, –1, 1) (ii) Find the length of the perpendicular from the point (1, 0, 1) to the plane x – y + z = 4. (a) 2 units (b) 4 units (c) 6 units (d) 8 units 3 3 3 3 (iii) The equation of the perpendicular from the point (1, 0, 1) to the plane x – y + z = 4 is (a) x − 1 = y+3 = z + 5 (b) x−1 = y+3 = z − 5 2 −1 2 −2 −1 2 (c) x − 1 = y = z − 1 (d) x − 1 = y = z − 1 1 −1 1 2 −2 1 (iv) The equation of the plane parallel to the plane x – y + z = 4, which is at a unit distance from the point (1, 0, 1) is (a) x – y + z + (2 − 3) (b) x – y + z – (2 + 3) (c) x – y + z + (2 + 3) (d) Both (a) and (c) (v) The direction cosine of the normal to the plane x – y + z = 4 is (a) ⎛ 1 , −1 , −1 ⎞ (b) ⎛ 1 , −1 , 1⎞ ⎝⎜ 3 3 3 ⎟⎠ ⎜⎝ 3 3 3 ⎟⎠ (c) ⎛ −1 , −1 , 1⎞ (d) ⎛ −1 , −1 , −1 ⎞ ⎝⎜ 3 3 3 ⎠⎟ ⎜⎝ 3 3 3 ⎠⎟ 11. A mobile tower stands at the top of a hill. Consider the surface on which the tower stands as a plane having points A (1, 0, 2), B (3, –1, 1) and C (1, 2, 1) on it. The mobile tower is tied with 3 cables from the point A, B and C such that it stands vertically on the ground. The top of the tower is at the point (2, 3, 1) as shown in the figure.