Question

The co-ordinates of the point where the line

$\frac{x - 6}{- 1} = \frac{y + 1}{0} = \frac{z + 3}{4}$ meets the plane $x + y - z = 3$are

• A. (2, 1, 0)
• B. (7, –1, –7)
• C. (1, 2, –6)
• D. (5, –1, 1)

Answer 1

Answer: (D)

(5, –1, 1)

Answer 2

Point on the line, $\frac { x - 6 } { - 1 } = \frac { y + 1 } { 0 } = \frac { z + 3 } { 4 } = r$ are

$( - r + 6 , - 1,4 r - 3 )$ This will be satisfy plane $x + y - z = 3$

$\therefore - r + 6 - 1 - 4 r + 3 = 3 \Rightarrow - 5 r + 5 = 0$ $\Rightarrow r = 1$

Required co-ordinates of point $\equiv ( 5 , - 1,1 )$.