Question

If $P(3,2,6)$ is a point in space and Q is a point on the line $r=(i-j+2k)+\mu (-3i+j+5k)$. Then the value of $\mu$for which the vector $PQ$ 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

The correct option is A

$\frac{1}{4}$

Explanation for the correct option

Any point on the line $r$ can be taken as, $\mathrm{Q}\equiv \left\{\right(1-3\mu ),(\mu -1),(5\mu +2\left)\right\}$.
So, $\overrightarrow{\mathrm{PQ}}=\{-3\mu -2,\mu -3,5\mu -4\}$
Since $\overrightarrow{\mathrm{PQ}}$ is parallel to the plane then it must be perpendicular to the normal vector of the plane.
$\begin{array}{rcl}1(-3\mu -2)-4(\mu -3)+3(5\mu -4)& =& 0\\ -3\mu -2-4\mu +12+15\mu -12& =& 0\\ 8\mu & =& 2\\ \mu & =& \frac{1}{4}\end{array}$

So, for value of $\mu =\frac{1}{4}$the vector $PQ$parallel to plane $x-4y+3z=1$.

Hence, option A is correct.