Question

The angle between the lines whose direction cosines satisfy the equations $l+m+n=0$ and $l^2+m^2+n^2$ is

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

Answer 1

Answer: (B)

$\frac{\pi }{3}$

According to question.................

$\begin{array}{c}\Rightarrow l+m+n=0----\left(i\right)\\ l=-(m+n)\\ \Rightarrow l^2=m^2+n^2---\left(ii\right)\\ \Rightarrow {\left(-(m+n)\right)}^2=m^2+n^2\\ \Rightarrow (m+n{)}^2=m^2+n^2\\ \Rightarrow m^2+n^2+2mn=m^2+n^2\\ \Rightarrow 2mn=0\\ \therefore mn=0\\ therearetwocase:m=0,n=0\\ case1:(m=0)\\ l=-n|usingequ:l+m+n=0----\left(i\right)\\ suppose:\\ l=k,m=o,n=-k\\ \Rightarrow l^2+m^2+n^2=1\\ \Rightarrow k^2+0+k^2=1\\ \Rightarrow k=\frac{1}{\sqrt{2}}\\ Apply,\\ \Rightarrow l=\frac{1}{\sqrt{2}},m=o,n=\frac{-1}{\sqrt{2}}(weAsume:l_1=\frac{1}{\sqrt{2}},m_1=o,n_1=\frac{-1}{\sqrt{2}}\\ case2:(n=0)\\ l=-m|usingequ:l+m+n=0----\left(i\right)\\ suppose:\\ \Rightarrow l=k,m=-k,n=0(\\ then,\\ l^2+m^2+n^2=1\\ k^2+k^2+0=1\\ k=\frac{1}{\sqrt{2}}\\ apply,\\ \Rightarrow l=\frac{1}{\sqrt{2}},m=\frac{-1}{\sqrt{2}},n=0(weAssume:l_2=\frac{1}{\sqrt{2}},m_2=\frac{-1}{\sqrt{2}},n_2=0\\ Anglebetween2lines=\theta \\ \cos \theta =\frac{(l_1+m_1+n_1).(l_2+m_2+n_2)}{\sqrt{({l_1}^2+{m_1}^2+{n_1}^2).({l_2}^2+{m_2}^2+{n_2}^2)}}\\ =\frac{l_1{l}_2+m_1{m}_2+n_1{n}_2}{\sqrt{({l_1}^2+{m_1}^2+{n_1}^2)({l_2}^2+{m_2}^2+{n_2}^2)}}\\ =\frac{\frac{1}{\sqrt{2}}.\frac{1}{\sqrt{2}}+0.\frac{-1}{\sqrt{2}}+\frac{-1}{\sqrt{2}}.0}{\sqrt{\left(\frac{1}{2}+\frac{1}{2}+0\right)\left(\frac{1}{2}+\frac{1}{2}+0\right)}}=\frac{\frac{1}{2}}{\sqrt{1}}\\ \cos \theta =\frac{1}{2},\therefore \theta =\frac{\pi }{3}\end{array}$

So,that the correct option is B.