Question

Find the angle between the two straight lines whose direction cosines $l,m,n$ are given by $2l+2m-n=0$ and $mn+nl+lm=0$

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{2}$

Answer 1

Answer: (D)

$\frac{\pi }{2}$

$2l+2m-n=0$ ...(1)

$mn+nl+lm=0$ ...(2)

Put $n=2l+2m$ in (2)

$\Rightarrow m(2l+2m)+l(2l+2m)+lm=0\Rightarrow 2l^2+5lm+2m^2=0$

$\Rightarrow (l+2m)(2l+m)=0$

Case 1: $l+2m-0\Rightarrow \frac{l}{-2}=\frac{m}{1}$

$\therefore$ from (1) $n=l\Rightarrow \frac{l}{-2}=\frac{n}{-2}\Rightarrow \frac{l}{-2}=\frac{m}{1}=\frac{n}{-2}$

$\therefore l_1=-2,m_1=1,n_1=-2$

Case 2: $2l+m=0\Rightarrow \frac{l}{1}=\frac{m}{-2}$

$\therefore$ from (1) $\frac{n}{-2}=\frac{m}{-2}\Rightarrow \frac{l}{1}=\frac{m}{-2}=\frac{n}{-2}$

$\Rightarrow l_1=1,m_2=-2,n_2=-2$

$\therefore l_1{l}_2+m_1{m}_2+n_1{n}_2=0\Rightarrow \theta =\frac{\pi }{2}$