Question

Image point of $(1,3,4)$ in the plane $2x - y + z + 3 = 0$ is

• A. (– 3, 5, 2)
• B. (3, 5, – 2)
• C. (3, – 5, 3)
• D. None of these

Answer 1

Answer: (A)

(– 3, 5, 2)

Answer 2

Let Q be image of the point $P ( 1,3,4 )$ in the given plane, then PQ is normal to the plane.

The d.r.'s of PQ are 2, –1,1

Since PQ passes through (1, 3, 4) and has d.r's 2, 1, –1;

therefore, equation of plane is $\frac { x - 1 } { 2 } = \frac { y - 3 } { - 1 } = \frac { z - 4 } { 1 } = r$, (say)

$\therefore$ $x = 2 r + 1 , y = - r + 3 , z = r + 4$

So, co-ordinates of Q be

$( 2 r + 1 , - r + 3 , r + 4 )$

Let R be the mid point of PQ, then co-ordiantes of R are $\left( \frac { 2 r + 1 + 1 } { 2 } , \frac { - r + 3 + 3 } { 2 } , \frac { r + 4 + 4 } { 2 } \right)$i.e., $\left( r + 1 , \frac { - r + 6 } { 2 } , \frac { r + 8 } { 2 } \right)$

Since R lies on the plane

$\therefore$ $2 ( r + 1 ) - \left( \frac { - r + 6 } { 2 } \right) + \left( \frac { r + 8 } { 2 } \right) + 3 = 0$ ⇒ $r = - 2$

So, co-ordinates of Q are (–3, 5, 2).

Trick : From option (1), mid point of (–3, 5, 2) and (1,3,4) satisfies the equation of plane $2 x - y + z + 3 = 0$ .