Mathematics - 3D-Coordinate Geometry Question with Solution | TestHub

The position vector of point $Q$ on the given line is $\vec{r}=(1-3\mu )\hat{i}+(-1+\mu )\hat{j}+(2+5\mu )\hat{k}$.

The position vector of point $P$ is $3\hat{i}+2\hat{j}+6\hat{k}$.

The vector $\overrightarrow{PQ}$ is given by the difference of the position vectors of $Q$ and $P$:

$\overrightarrow{PQ}=(1-3\mu -3)\hat{i}+(-1+\mu -2)\hat{j}+(2+5\mu -6)\hat{k}$

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

Since the vector $\overrightarrow{PQ}$ is parallel to the plane $x-4y+3z=1$, it must be perpendicular to the normal vector of the plane, which is $\vec{n}=\hat{i}-4\hat{j}+3\hat{k}$.

Therefore, the dot product $\overrightarrow{PQ}\cdot \vec{n}=0$:

$(-2-3\mu )(1)+(-3+\mu )(-4)+(-4+5\mu )(3)=0$

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

$8\mu -2=0$

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

Answer: $\frac{1}{4}$