Question

The acute angle between the lines whose direction cosines satisfy \[ l^2-5m^2+n^2=0 \] and \[ l+m-n=0 \] is

• A. \[ \cos^{-1}\!\left(\frac{\sqrt3}{4}\right) \]
• B. \[ \frac{\pi}{3} \]
• C. \[ \cos^{-1}\!\left(\frac23\right) \]
• D. \[ \frac{\pi}{6} \] \bigskip

Hint:

When direction cosines satisfy two equations, first eliminate one variable to obtain the possible direction ratios. Each factor gives a distinct line. Then use the dot-product formula to find the angle between them.

Answer 1

The Correct Option is B

Concept: Direction cosines satisfy \(l^2+m^2+n^2=1\). Use the two given linear relations to find possible direction ratios for the two lines, then compute the acute angle via dot product.

Step 1: From \(l+m-n=0 \Rightarrow n=l+m\). Substitute into \(l^2-5m^2+n^2=0\): \(l^2-5m^2+(l+m)^2=0 \Rightarrow 2l^2+2lm-4m^2=0 \Rightarrow l^2+lm-2m^2=0 \Rightarrow (l-m)(l+2m)=0\).

Step 2: Case 1: \(l=m \Rightarrow n=2l\), ratios (1,1,2). Case 2: \(l=-2m \Rightarrow n=-m\), ratios (-2,1,-1).

Step 3: Vectors \(\vec{d}_1=(1,1,2), \vec{d}_2=(-2,1,-1)\). Dot = -2+1-2 = -3. Magnitudes: \(|\vec{d}_1|=\sqrt6, |\vec{d}_2|=\sqrt6\). \(\cos\theta = \frac{|-3|}{6} = \frac12 \Rightarrow \theta = \pi/3\).

Step 4: Write the final answer. \(\boxed{\frac{\pi}{3}}\)