Higher Engineering Mathematics - BS Grewal

284 Higher Engineering Mathematics magnitude, direction and sense), then AP = λb, where λ Hence, the Cartesian equations are: is a scalar quantity. Hence, from above, r=a+λb x −2 = y −3 = z − (−1) = λ 1 −2 3 (8) If, say, r = xi + yj + zk, a =a1i +a2 j + a3k and i.e. x −2 = 3 − y = z + 1 = λ b = b1i + b2 j + b3k, then from equation (8), 23 xi + yj + zk = (a1i + a2 j + a3k) Problem 12. The equation + λ(b1i + b2 j + b3k) 2x − 1 = y + 4 = −z + 5 Hence x = a1 + λb1, y = a2 + λb2 and z = a3 + λb3. 33 2 Solving for λ gives: represents a straight line. Express this in vector form. x −a1 = y − a2 = z − a3 = λ (9) b1 b2 b3 Comparing the given equation with equation (9), shows Equation (9) is the standard Cartesian form for the vector that the coefficients of x, y and z need to be equal to equation of a straight line. unity. Problem 11. (a) Determine the vector equation of Thus 2x − 1 = y + 4 = −z + 5 becomes: the line through the point with position vector 33 2 2i + 3j −k which is parallel to the vector i − 2j + 3k. (b) Find the point on the line corresponding to λ =3 x − 1 = y +4 = z−5 in the resulting equation of part (a). 2 3 −2 (c) Express the vector equation of the line in standard Cartesian form. 3 (a) From equation (8), 2 r = a + λb Again, comparing with equation (9), shows that i.e. r = (2i + 3j −k) +λ(i − 2j + 3k) a1 = 1 , a2 = −4 and a3 = 5 and 2 or r = (2 + λ)i + (3 − 2λ)j + (3λ − 1)k 3 which is the vector equation of the line. b1 = 2 , b2 = 3 and b3 = −2 (b) When λ =3, r = 5i −3j + 8k. (c) From equation (9), In vector form the equation is: x − a1 = y − a2 = z − a3 = λ r = (a1 + λb1)i + (a2 + λb2) j + (a3 + λb3)k, b1 b2 b3 from equation (8) Since a = 2i + 3j − k, then a1 = 2, i.e. r = 1 + 3 λ i + (−4 + 3λ) j + (5 − 2λ)k 22 a2 = 3 and a3 = −1 and or r = 1 (1 + 3λ)i + (3λ − 4) j + (5 − 2λ)k b = i − 2j + 3k, then 2 b1 = 1, b2 = −2 and b3 = 3 Now try the following exercise Exercise 114 Further problems on the vector equation of a line 1. Find the vector equation of the line through the point with position vector 5i −2j + 3k which

Scalar and vector products 285 is parallel to the vector 2i + 7j −4k. Determine In problems 3 and 4, express the given straight line equations in vector form. the point on the line corresponding to λ =2 in the resulting equation. 3. 3x − 1 = 5y + 1 = 4 − z ⎡⎤ 4 23 r = (5 + 2λ)i + (7λ − 2)j ⎣ + (3 − 4λ)k; ⎦ r = 1 (1 + 4λ)i + 1 (2λ − 1)j 3 5 r = 9i + 12j − 5k + (4 − 3λ)k 2. Express the vector equation of the line in 4. 2x + 1 = 1 −4y = 3z −1 problem 1 in standard Cartesian form. 54 x−5 = y+2 = 3−z =λ r = 1 (λ − 1)i + 1 (1 − 5λ)j 2 74 2 4 + 1 (1 + 4λ)k 3