Question

The angle between two lines $\frac{x+1}{2}=\frac{y+3}{2}=\frac{z-4}{-1}$ and $\frac{x-4}{1}=\frac{y+4}{2}=\frac{z+1}{2}$ is :

• A. ${\cos }^{-1}\left(\frac{1}{9}\right)$
• B. ${\cos }^{-1}\left(\frac{4}{9}\right)$
• C. ${\cos }^{-1}\left(\frac{2}{9}\right)$
• D. ${\cos }^{-1}\left(\frac{3}{9}\right)$

Answer 1

Answer: (B)

Note: The angle $\theta$ between the two lines
$\frac{x-x_1}{a_1}=\frac{y-y_1}{a_2}=\frac{z-z_1}{a_3}$
and $\frac{x-x_2}{b_1}=\frac{y-y_2}{b_2}=\frac{z-z_2}{b_3}$ is given by:
$\cos \theta =\frac{a_1{b}_1+a_2{b}_2+a_3{b}_3}{\sqrt{a_1^2+a_2^2+a_3^2}\sqrt{b_1^2+b_2^2+b_3^2}}$
Now in the given equation:
$a_1=2,a_2=2,a_3=-1$
$b_1=1,b_2=2,b_3=2$
$\therefore \cos \theta =\frac{2\times 1+2\times 2+\left(-2\right)\times 1}{\sqrt{4+4+1}\sqrt{4+4+1}}=\frac{4}{9}$
$\Rightarrow \theta ={\cos }^{-1}\left(\frac{4}{9}\right)$