Question

The acute angle between two lines such that the direction cosines $l,m,n$ of each of them satisfy the equation $l+m+n=0$ and $l^2+m^2-n^2=0$ is

• A. ${30}^{\circ }$
• B. ${45}^{\circ }$
• C. ${60}^{\circ }$
• D. ${15}^{\circ }$

Answer 1

Answer: (C)

${60}^{\circ }$

Lines are $l+m+n=0⟹-l=(m+n)$and $l^2-m^2+n^2=0⟹l^2=m^2-n^2$Solving them gives,$(-(m+n){)}^2=m^2+n^2+2mn=m^2-n^2⟹2n(n+m)=0$Now, for $n=0$$m=-l$, so d.c's $(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}},0)$And for $n=-m$$l=0$, so d.c's $(0,\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}})$Angles between the lines $|\cos \theta |=\left|\frac{1}{2}\right|⟹\theta =\frac{\pi }{3}$