1)

The angle between the lines   $\frac{x-1}{4}=\frac{y-3}{1}=\frac{z}{8}$  and

$\frac{x-2}{2}=\frac{y+1}{2}=\frac{z-4}{1}$


A) $\cos^{-1}\left(\frac{2}{3}\right)$

B) $\cos^{-1}\left(\frac{1}{2}\right)$

C) $\cos^{-1}\left(\frac{3}{4}\right)$

D) $\cos^{-1}\left(\frac{1}{3}\right)$

Answer:

Option A

Explanation:

Given line,

$\frac{x-1}{4}=\frac{y-3}{1}=\frac{z}{8}$  and

$\frac{x-2}{2}=\frac{y+1}{2}=\frac{z-4}{1}$

 Here, a1=4, b1=1, c1=8

 a2=2, b2 =2, c2=1

$\therefore$    $\cos\theta=\frac{a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}}{(\sqrt{a_1^2+b_1^2+c_1^2})(\sqrt{a_2^2+b_2^2+c_2^2})}$

$\Rightarrow $        $\cos\theta=\frac{8+2+8}{(\sqrt{16+1+64})(\sqrt{4+4+1})}$

$\Rightarrow $       $\cos\theta=\frac{18}{9\times3}=\frac{2}{3}$

 $\Rightarrow $     $ \theta=\cos^{-1}\left(\frac{2}{3}\right)$