Question

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

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

Answer 1

Answer: (A)

$\frac{1}{4}$

$\overrightarrow{Q}=(1-3\mu )i+j(\mu -1)+k(2+5\mu )$

Therefore

$\overrightarrow{PQ}=(-2-3\mu )i+(\mu -3)j+(5\mu -4)k$

Now it is parallel to the lane.

Hence the direction vector is perpendicular to the normal of the plane.

Hence their dot products will be $0$.

Therefore

$-2-3\mu -4\mu +12+15\mu -12=0$

$8\mu -2=0$

$\mu =\frac{2}{8}=\frac{1}{4}$