Question

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

• A. $\frac{1}{2}$
• B. $\frac{1}{4}$
• C. $\frac{1}{8}$
• D. $-\frac{1}{4}$

Answer 1

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

The position vector of $A=\overrightarrow{OA}=(1-3\mu )\hat{i}+(\mu -1)\hat{j}+(2+5\mu )\hat{k}$
position vector of $B=\overrightarrow{OB}=3\hat{i}+2\hat{j}+6\hat{k}$
$\therefore \overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}$
$=(3\mu +2)\hat{i}+(3-\mu )\hat{j}+(4-5\mu )\hat{k}$
$\overrightarrow{AB}$ is parallel to the plane $x-4y+3z=1$
So, $(3\mu +2)1+(3-\mu )(-4)+(4-5\mu )3=0$
$\Rightarrow 3\mu +2-12+4\mu +12-15\mu =0$
$\Rightarrow -8\mu +2=0$
$\Rightarrow \mu =\frac{1}{4}$