Question

The angle between pair of lines whose direction-ratios are $1,1,2$ and $\sqrt{3}-1,-\sqrt{3}-1,4$ is

• A. ${30}^o$
• B. ${45}^o$
• C. ${60}^o$
• D. ${90}^o$

Answer 1

The correct option is C ${60}^o$

Angle between the pair of lines whose direction ratios are $a_1,b_1,c_1$ and $a_2,b_2,c_2$ is

$\cos \theta =\frac{|a_1{a}_2+b_1{b}_2+c_1{c}_2|}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}$

$a_1=1,b_1=1,c_1=2;a_2=\sqrt{3}-1,b_2=-\sqrt{3}-1,c_2=4$

$\therefore \cos \theta =\frac{|\sqrt{3}-1-\sqrt{3}-1+8|}{\sqrt{1^2+1^2+2^2}\sqrt{(\sqrt{3}-1{)}^2+(-\sqrt{3}-1{)}^2+4^2}}$

$\cos \theta =\frac{|6|}{\sqrt{6}\sqrt{3+1-2\sqrt{3}+3+1+2\sqrt{3}+16}}$

$\cos \theta =\frac{6\sqrt{6}}{\sqrt{6}\sqrt{24}}$

$\cos \theta =\frac{\sqrt{6}}{2\sqrt{6}}$

$\cos \theta =\frac{1}{2}$

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

$\therefore ={\cos }^{-1}\left(\cos {60}^o\right)$

$\therefore \theta ={60}^o$.