Question

The angle between the lines whose direction cosines satisfy the equations
\[ l+m+n=0,\quad l^2+m^2-n^2=0 \] is

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

Hint:

When direction cosines satisfy equations, solve for possible direction ratios. Two solutions give two lines, then use dot product for angle.

Answer 1

The Correct Option is C

Step 1: Solve equations for direction ratios.
Given:
\[ l+m+n=0 \Rightarrow n=-(l+m) \]
Second equation:
\[ l^2+m^2-n^2=0 \Rightarrow l^2+m^2=(l+m)^2 \]
\[ l^2+m^2=l^2+m^2+2lm \Rightarrow 2lm=0 \Rightarrow lm=0 \]
Step 2: Cases.
Either \(l=0\) or \(m=0\).
Case 1: \(l=0\). Then \(n=-(0+m)=-m\).
Direction ratios \((0,1,-1)\).
Case 2: \(m=0\). Then \(n=-(l+0)=-l\).
Direction ratios \((1,0,-1)\).
Step 3: Find angle between these two lines.
Let vectors:
\[ \vec{a}=(0,1,-1),\quad \vec{b}=(1,0,-1) \]
Dot product:
\[ \vec{a}\cdot\vec{b}=0\cdot 1+1\cdot 0+(-1)(-1)=1 \]
Magnitudes:
\[ |\vec{a}|=\sqrt{0^2+1^2+(-1)^2}=\sqrt{2} \]
\[ |\vec{b}|=\sqrt{1^2+0^2+(-1)^2}=\sqrt{2} \]
\[ \cos\theta = \frac{1}{2} \Rightarrow \theta=\frac{\pi}{3} \]
Final Answer:
\[ \boxed{\frac{\pi}{3}} \]