Question

The image of the point $3\hat{i}-2\hat{j}+\hat{k}$ in the plane $\overline{r}.(3\hat{i}-\hat{j}+4\hat{k})=2$

• A. $-\hat{j}+3\hat{k}$
• B. $\hat{j}-3\hat{k}$
• C. $-\hat{j}-3\hat{k}$
• D. $-2\hat{j}-3\hat{k}$

Answer 1

The correct option is C $-\hat{j}-3\hat{k}$

$PQ$ is normal to the plane therefore $\overline{PQ}\parallel 3\hat{i}-\hat{j}+4\hat{k}$
As $PQ$ passes through $3\hat{i}-2\hat{j}+\hat{k}$
$\therefore$ Equations of lines $PQ$ is given by
$\overline{r}=(3\hat{i}-2\hat{j}+\hat{k})+\lambda (3\hat{i}-\hat{j}+4\hat{k})$
$=(3+3\lambda )\hat{i}+\hat{j}(-2-\lambda )+\hat{k}(4\lambda +1)$
$\therefore$ General point lies on the line is
$(3+3\lambda )\hat{i}+\hat{j}(-2-\lambda )+\hat{k}(4\lambda +1)$
Now general point lies on the line
$\therefore (3+3\lambda )\left(3\right)+(-1)(-2-\lambda )+4(4\lambda +1)-2=0$
$\therefore \lambda =-\frac{1}{2}$
Now $M$ is mid point of $PQ$ where $M(\frac{3}{2},\frac{-3}{2},-1)$
$\therefore Q(x{}^{\prime },y{}^{\prime },z{}^{\prime })=(0,-1,-3)$
$\therefore$ Image of point $P$ is $-\hat{j}-3\hat{k}$
Hence (c) is the correct answer.