If the two lines l_1: x-23=y+1-2, z=2 and l_2: x-11=2y+3=z+52 are perpendicular, then an a…

Question

If the two lines $l_1:\ \dfrac{x-2}{3}=\dfrac{y+1}{-2},\ z=2$ and $l_2:\ \dfrac{x-1}{1}=\dfrac{2y+3}{\alpha}=\dfrac{z+5}{2}$ are perpendicular, then an angle between the lines $l_2$ and $l_3:\ \dfrac{1-x}{3}=\dfrac{2y-1}{-4}=\dfrac{z}{4}$ is:

Accepted Answer

$l_1$ direction $(3,-2,0)$. $l_2$: rewrite $\dfrac{2y+3}{\alpha}=\dfrac{y+3/2}{\alpha/2}$, direction $\left(1,\dfrac{\alpha}{2},2\right)$.

Perpendicularity: $3(1)-2\left(\dfrac{\alpha}{2}\right)=0 \implies \alpha=3$, so $l_2$ direction (scaled) $=(2,3,4)$.

$l_3$: rewrite as $\dfrac{x-1}{-3}=\dfrac{y-\frac12}{-2}=\dfrac{z}{4}$, direction $(-3,-2,4)$.
$$\cos\theta=\frac{(2)(-3)+(3)(-2)+(4)(4)}{\sqrt{29}\sqrt{29}}=\frac{4}{29}.$$
$$\theta=\cos^{-1}\left(\frac{4}{29}\right)=\sec^{-1}\left(\frac{29}{4}\right).$$

Solutions · 1