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

The position vector of point $P$ is $2\hat{i}+\hat{j}+5\hat{k}$.

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

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

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

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

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

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

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

$-3-9\mu -\mu +2+20\mu -12=0$

$10\mu -13=0$

$\mu =\frac{13}{10}$

Answer: $\frac{13}{10}$