Question

The image of the point having the position vector in the plane $\overrightarrow { \mathbf { r } }$ . + 3 = 0 is

• A. $3\widehat{i} - 5\widehat{j} + 2\widehat{k}$
• B. $3\widehat{i} - 5\widehat{j} + 4\widehat{k}$
• C. $- 3\widehat{i} + 5\widehat{j} + 2\widehat{k}$
• D. $3\widehat{i} + 5\widehat{j} + 4\widehat{k}$

Answer 1

Answer: (C)

$- 3\widehat{i} + 5\widehat{j} + 2\widehat{k}$

Answer 2

Equation of line passing through given point P and normal to given plane is

= $( \hat { \mathrm { i } } + 3 \hat { \mathrm { j } } + 4 \hat { \mathrm { k } } )$ + l

Ž Image point Q

ŗ$( ( 1 + 2 \lambda ) \hat { i } + ( 3 - \lambda ) \hat { j } + ( 4 + \lambda ) \hat { k } )$

Now mid point of PQ ŗ

satisfies given plane that gives l = –2

So position vector of image is