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

A) $\frac{1}{4}$
B) $- \frac{1}{4}$
C) $\frac{1}{8}$
D) $- \frac{1}{8}$
Explanation

(a)
Given,
$\overrightarrow{\mathbf{OQ}} = (1 - 3\mu)\widehat{\dot{\mathbf{i}}} + (\mu - 1)\widehat{\dot{\mathbf{j}}} + (5\mu + 2)\widehat{\mathbf{k}}$
$\overrightarrow{\mathbf{OP}} = 3\widehat{\dot{\mathbf{i}}} + 2\widehat{\dot{\mathbf{j}}} + 6\widehat{\mathbf{k}}$ (where $O$ is origin)
[image:10_image15.emf]
Now,
$\ \overrightarrow{\mathbf{PQ}} = (1 - 3\mu - 3)\widehat{\dot{\mathbf{i}}} + (\mu - 1 - 2)\widehat{\dot{\mathbf{j}}} + (5\mu + 2 - 6)\widehat{\mathbf{k}}$
$= ( - 2 - 3\mu)\widehat{\dot{\mathbf{i}}} + (\mu - 3)\widehat{\dot{\mathbf{j}}} + (5\mu - 4)\widehat{\mathbf{k}}$
$\because\overrightarrow{\mathbf{PQ}}$ is parallel to the plane $x - 4y + 3z = 1\$
$\therefore\ - 2 - 3\mu - 4\mu + 12 + 15\mu - 12 = 0$
$\Longrightarrow 8\mu = 2$
$\Longrightarrow \ \mu = \frac{1}{4}$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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