Question

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

• A. ${\cos }^{-1}\left(\frac{29}{4}\right)$
• B. ${\sec }^{-1}\left(\frac{29}{4}\right)$
• C. ${\cos }^{-1}\left(\frac{2}{29}\right)$
• D. ${\cos }^{-1}\left(\frac{2}{\sqrt{29}}\right)$

Answer 1

Answer: (B)

$l_1:\frac{x-2}{3}=\frac{y+1}{-2}=\frac{z-2}{0}$
$l_2:\frac{x-1}{1}=\frac{y+3/2}{\alpha /2}=\frac{z+5}{2}$
$l_3:\frac{x-1}{-3}=\frac{y-1/2}{-2}=\frac{z-0}{4}$
$l_1\perp l_2\Rightarrow \frac{|3-\alpha +0|}{\sqrt{13}\sqrt{1+\frac{{\alpha }^2}{4}+4}}=0\Rightarrow \alpha =3$
angle between $l_2\&l_3$
$\cos \theta =\frac{|1\times (-3)+(-2)(\alpha /2)+2\times 4|}{\sqrt{1+4+\frac{{\alpha }^2}{4}}\sqrt{9+16+4}}$
$\cos \theta =\frac{|-3-\alpha +8|}{\sqrt{5+\frac{{\alpha }^2}{4}}\sqrt{29}}$
$\mathrm{put}\alpha =3$
$\cos \theta =\frac{2}{\sqrt{\frac{29}{4}}\sqrt{29}}=\frac{4}{29}$
$\theta ={\cos }^{-1}\left(\frac{4}{29}\right)\Rightarrow \theta ={\sec }^{-1}\left(\frac{29}{4}\right)$