Question

The acute angle between the lines $x=-2+2t,y=3-4t,z=-4+t$ and $x=2-t,y=3+2t,z=-4+3t$ is

• A. ${\sin }^{-1}\frac{1}{\sqrt{3}}$
• B. ${\cos }^{-1}\frac{1}{\sqrt{6}}$
• C. ${\cos }^{-1}\frac{1}{\sqrt{5}}$
• D. ${\cos }^{-1}2/3$

Answer 1

Answer: (B)

${\cos }^{-1}\frac{1}{\sqrt{6}}$

Given lines are $x=-2+2t,y=3-4t,z=-4+t$ and $x=2-t,y=3+2t,z=-4+3t$
$\Rightarrow \frac{x+2}{2}=\frac{y-3}{-4}=\frac{z+4}{1}$ and $\frac{x-2}{-1}=\frac{y-3}{2}=\frac{z+4}{3}$
if $a_1,b_1,c_1$ and $a_2,b_2,c_2$ are direction ratios of two lines, then angle between them is
$\cos \theta =\frac{a_1{a}_2+b_1{b}_2+c_1{c}_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}$
$\Rightarrow \cos \theta =\frac{-1}{\sqrt{6}}$
$\Rightarrow \theta =\pi -{\cos }^{-1}\left(\frac{1}{\sqrt{6}}\right)$
Hence, acute angle between the lines is ${\cos }^{-1}\left(\frac{1}{\sqrt{6}}\right)$