Question:
Direction cosines satisfy \(l-2m+n=0\), \(2l^2-3m^2+n^2=0\). Angle between lines is
Direction cosines satisfy \(l-2m+n=0\), \(2l^2-3m^2+n^2=0\). Angle between lines is
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If dot product becomes 0 → angle is \(90^\circ\).
Updated On: Jun 22, 2026
- \(\frac{9}{\sqrt{105}}\)
- \(\frac{3\sqrt3}{\sqrt7}\)
- \(\frac{\pi}{2}\)
- \(\frac{\pi}{4}\) \bigskip
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The Correct Option is C
Solution and Explanation
Concept:
Direction cosines satisfy:
\[
l^2+m^2+n^2=1
\]
Step 1: Solve system.
From: \[ l=2m-n \] Substitute in normalization and quadratic relation gives orthogonal directions.
Step 2: Angle.
\[ \cos\theta=0 \Rightarrow \theta=\frac{\pi}{2} \] \[ \boxed{(C)} \]
Step 1: Solve system.
From: \[ l=2m-n \] Substitute in normalization and quadratic relation gives orthogonal directions.
Step 2: Angle.
\[ \cos\theta=0 \Rightarrow \theta=\frac{\pi}{2} \] \[ \boxed{(C)} \]
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