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 \[ l^2+m^2+n^2=1. \] The given relations determine two possible lines. The acute angle between them is obtained from the dot product of their direction ratios.

Step 1: Use the relation \(l+m-n=0\). \[ n=l+m. \] Substitute into \[ l^2-5m^2+n^2=0. \] \[ l^2-5m^2+(l+m)^2=0. \] \[ 2l^2+2lm-4m^2=0. \] \[ l^2+lm-2m^2=0. \] \[ (l-m)(l+2m)=0. \] Hence, \[ l=m \] or \[ l=-2m. \]

Step 2: Find the two lines. Case 1: \[ l=m. \] Then \[ n=l+m=2l. \] Direction ratios are \[ (1,1,2). \] Case 2: \[ l=-2m. \] Then \[ n=l+m=-m. \] Direction ratios are \[ (-2,1,-1). \] Thus the two lines have direction vectors \[ \vec d_1=(1,1,2), \qquad \vec d_2=(-2,1,-1). \]

Step 3: Find the angle between the lines. \[ \vec d_1\cdot\vec d_2 = (1)(-2)+(1)(1)+(2)(-1) = -3. \] For the acute angle, \[ \cos\theta = \frac{|\vec d_1\cdot\vec d_2|} {|\vec d_1||\vec d_2|}. \] \[ = \frac{3} {\sqrt{1+1+4}\,\sqrt{4+1+1}}. \] \[ = \frac{3}{\sqrt6\cdot\sqrt6}. \] \[ = \frac12. \] Hence, \[ \theta=\cos^{-1}\!\left(\frac12\right). \] \[ \theta=\frac{\pi}{3}. \]

Step 4: Write the final answer. \[ \boxed{\frac{\pi}{3}} \]