Question

Let the foot of perpendicular from a point $P(1,2,–1)$ to the straight line $L:\frac{x}{1}=\frac{y}{0}=\frac{z}{-1}$ be $N$. Let a line be drawn from $P$ parallel to the plane $x+y+2z=0$ which meets $L$ at point $Q$. If $\alpha$ is the acute angle between the lines $PN$ and $PQ,$ then $\cos \alpha$ is equal to

• A. $\frac{\sqrt{3}}{2}$
• B. $\frac{1}{2\sqrt{3}}$
• C. $\frac{1}{\sqrt{3}}$
• D. $\frac{1}{\sqrt{5}}$

Answer 1

The correct option is C $\frac{1}{\sqrt{3}}$


$L:\frac{x}{1}=\frac{y}{0}=\frac{z}{-1}=r$
$N\equiv (r_1,0,–r_1)$
$P(1,2,–1)$
D.R's of $NP=(r_1–1,–2,–r_1+1)\perp \text{with D.R's}(1,0,–1)$
$\Rightarrow r_1-1+0+r_1-1=0$
$\Rightarrow r_1=1$
$\Rightarrow N(1,0,-1)$
$Q(r_2,0,–r_2)$
D.R's of normal of plane $x+y+2z=0$ are $(1,1,2)$
D.R's of $PQ=(r_2–1,–2,–r_2+1)$ are perpendicular to $(1,1,2)$
$\Rightarrow r_2-1-2-2r_2+2=0$
$\Rightarrow r_2=-1$
$\Rightarrow Q(-1,0,1)$
$NP=2,PQ=2\sqrt{3}$
$\cos \alpha =\frac{NP}{PQ}=\frac{1}{\sqrt{3}}$