Question

The angle between the lines $\frac{x+2}{1}=\frac{y+3}{-2}=\frac{z-4}{1}$ and $\frac{x+1}{-1}=\frac{y}{1}=\frac{z}{0}$ is

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{2}$
• D. $0$

Answer 1

Answer: (A)

$\frac{\pi }{6}$

Angle between the lines $l_1=\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}$ and $l_2=\frac{x-x^2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}$ is given by

$\cos \theta =\frac{|a_1\cdot a_2+b_1\cdot b_2+c_1\cdot c_2|}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}$

$\therefore$ Angle between the lines $\frac{x+2}{1}=\frac{y+3}{-2}=\frac{z-4}{1}$ and $\frac{x+1}{-1}=\frac{y}{1}=\frac{z}{0}$ is

$\cos \theta =\frac{|1(-1)+(-2)\cdot 1+1\left(0\right)|}{\sqrt{(1{)}^2+(-2{)}^2+(1{)}^2}\sqrt{(-1{)}^2+(1{)}^2+(0{)}^2}}$

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

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

$\cos \theta =\left|\frac{-\sqrt{3}}{2}\right|$

$\theta ={\cos }^{-1}\left(\frac{-\sqrt{3}}{2}\right)={\cos }^{-1}\left(\frac{\sqrt{3}}{2}\right)=\frac{\pi }{6}$

$\therefore \theta =\frac{\pi }{6}$.