Question

If the two lines $l_{1}: \frac{ x -2}{3}=\frac{ y +1}{-2}, z =2$ and $l_{2}: \frac{ x -1}{1}=\frac{2 y +3}{\alpha}=\frac{ z +5}{2}$ perpendicular, then an angle between the lines $l_{2}$ and $l_{3}: \frac{1- x }{3}=\frac{2 y -1}{-4}=\frac{ z }{4}$ 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 $l_{2} \& l_{3}$
$\cos \theta=\frac{|1 \times(-3)+(-2)(\alpha / 2)+2 \times 4|}{\sqrt{1+4+\frac{\alpha^{2}}{4}} \sqrt{9+16+4}}$
$\cos \theta=\frac{|-3-\alpha+8|}{\sqrt{5+\frac{\alpha^{2}}{4}} \sqrt{29}}$
$\operatorname{put} \alpha=3$
$\cos \theta=\frac{2}{\sqrt{\frac{29}{4}} \sqrt{29}}=\frac{4}{29}$
$\theta=\cos ^{-1}\left(\frac{4}{29}\right) \Rightarrow \theta=\sec ^{-1}\left(\frac{29}{4}\right)$