# Let the foot of perpendicular from a point

Question:

Let the foot of perpendicular from a point $\mathrm{P}(1,2,-1)$ to the straight line $\mathrm{L}: \frac{\mathrm{x}}{1}=\frac{\mathrm{y}}{0}=\frac{\mathrm{z}}{-1}$ be $\mathrm{N}$.

Let a line be drawn from $P$ parallel to the plane $x+y+2 z=0$ which meets $L$ at point $Q$. If $\alpha$ is the acute angle between the lines $\mathrm{PN}$ and $\mathrm{PQ}$, then $\cos \alpha$ is equal to_________.

1. $\frac{1}{\sqrt{5}}$

2. $\frac{\sqrt{3}}{2}$

3. $\frac{1}{\sqrt{3}}$

4. $\frac{1}{2 \sqrt{3}}$

Correct Option: , 3

Solution:

$\overrightarrow{P N} \cdot(\hat{i}-\hat{k})=0$

$\Rightarrow \mathrm{N}(1,0,-1)$

Now,

$\overrightarrow{\mathrm{PQ}} \cdot(\hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}})=0$

$\Rightarrow \mu=-1$

$\Rightarrow Q(-1,0,1)$

$\overrightarrow{\mathrm{PN}}=2 \hat{\mathrm{j}}$ and $\overrightarrow{\mathrm{PQ}}=2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}$

$\Rightarrow \cos \alpha=\frac{1}{\sqrt{3}}$