Question

Let $P(3,2,6)$ be a point in space and $Q$ be a point on the line $\overrightarrow{r}=\hat{i}-\hat{j}+2\hat{k}+s(-3\hat{i}+\hat{j}+5\hat{k})$. Then the value of $s$ for which $\overrightarrow{PQ}$ is parallel to the plane $x-4y+3z=1$

• A. $\frac{1}{4}$
• B. $\frac{-1}{4}$
• C. $\frac{1}{8}$
• D. $\frac{-1}{8}$

Answer 1

The correct option is A $\frac{1}{4}$

Position vector of $P$ is $3\hat{i}+2\hat{j}+6\hat{k}$
$\overrightarrow{PQ}=\overrightarrow{r}-(3\hat{i}+2\hat{j}+6\hat{k})$
$=(\hat{i}-\hat{j}+2\hat{k})+s(-3\hat{i}+\hat{j}+5\hat{k})-(3\hat{i}+2\hat{j}+6\hat{k})$
$=(-3s-2)\hat{i}+(s-3)\hat{j}+(5s-4)\hat{k}$
$\overrightarrow{PQ}.(\hat{i}-4\hat{j}+3\hat{k})=0$
$\Rightarrow [-(3s+2)\hat{i}+(s-3)\hat{j}+(5s-4)\hat{k}].(\hat{i}-4\hat{j}+3\hat{k})=0$
$\Rightarrow -3s-2-4s+12+15s-12=0$
$\Rightarrow 8s-2=0$
$\Rightarrow 8s=2$
$\therefore s=\frac{2}{8}=\frac{1}{4}$