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 \times1 +2\times 2+\left(-2\right)\times 1}{\sqrt{4+4+1}\sqrt{4+4+1}} = \frac{4}{9} $
$\Rightarrow \theta = \cos^{-1} \left( \frac{4}{9}\right)$