Question

The position vector of a point in which a line through the origin and perpendicular to the plane $2x-y-z=4,$ meets the plane $\overrightarrow{r}\cdot (3\hat{i}-5\hat{j}+2\hat{k})=6$ is

• A. $(1,-1,-1)$
• B. $(-1,-1,2)$
• C. $(4,2,2)$
• D. $(\frac{4}{3},\frac{-2}{3},\frac{-2}{3})$

Answer 1

The correct option is D $(\frac{4}{3},\frac{-2}{3},\frac{-2}{3})$

Vector perpendicular to $2x-y-z=4$ is $\overrightarrow{n}=2\hat{i}-\hat{j}-\hat{k}$
Equation of line through origin and parallel to $\overrightarrow{n}=2\hat{i}-\hat{j}-\hat{k}$ is $\overrightarrow{r}=\lambda (2\hat{i}-\hat{j}-\hat{k})$
This line intersects the plane $\overrightarrow{r}\cdot (3\hat{i}-5\hat{j}+2\hat{k})=6$
$\Rightarrow \lambda (6+5-2)=6$
$\Rightarrow \lambda =\frac{2}{3}$
$\therefore$ Position vector of intersection point is
$\overrightarrow{r}=\frac{2}{3}(2\hat{i}-\hat{j}-\hat{k})=\frac{4}{3}\hat{i}-\frac{2}{3}\hat{j}-\frac{2}{3}\hat{k}$