Let the foot of perpendicular from a point P(1,2,-1) to the straight line L: x1=y0=z-1 be …

Question

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

Accepted Answer

$$\overrightarrow {PN} .(\widehat i - \widehat k) = 0$$

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

Now,

$$\overrightarrow {PQ} .(\widehat i + \widehat j + 2\widehat k) = 0$$

$$\Rightarrow$$ $$\mu$$ = - 1

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

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

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

Solutions · 1