Question

The line joining the points (3, 5, –7) and (–2, 1, 8) meets the yz-plane at point

• A. $\left( 0,\frac{13}{5},2 \right)$
• B. $\left( 2,0,\frac{13}{5} \right)$
• C. $\left( 0,2,\frac{13}{5} \right)$
• D. (2, 2, 0)

Answer 1

Answer: (A)

$\left( 0,\frac{13}{5},2 \right)$

Answer 2

Line joining the points (3,5,–7) and (–2,1,8) is,

$\frac { x - 3 } { ( - 2 ) - ( 3 ) } = \frac { y - 5 } { ( 1 ) - ( 5 ) } = \frac { z - ( - 7 ) } { 8 - ( - 7 ) }$

$\frac { x - 3 } { - 5 } = \frac { y - 5 } { - 4 } = \frac { z + 7 } { 15 } = K$, (Let) …..(i)

∴ $x = - 5 K + 3$, $y = - 4 K + 5$, $z = 15 K - 7$

$\bullet \bullet$ Line (i) meets the yz-plane

$\therefore$ $- 5 K + 3 = 0 \Rightarrow K = 3 / 5$

Put the value of K in $x , y , z$

So the required point is (0, 13/5, 2).