Question:
The angle between the lines \( \frac{x-3}{1} = \frac{y+1}{-1} = \frac{z-2}{-1} \) and \( \frac{x+1}{2} = \frac{y-2}{2} = \frac{z+3}{-2} \) is
The angle between the lines \( \frac{x-3}{1} = \frac{y+1}{-1} = \frac{z-2}{-1} \) and \( \frac{x+1}{2} = \frac{y-2}{2} = \frac{z+3}{-2} \) is
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Extract direction ratios directly from symmetric form of line equations.
Updated On: Apr 21, 2026
- \( \cos^{-1}\left(\sqrt{\frac{2}{6}}\right) \)
- \( \cos^{-1}\left(\sqrt{\frac{6}{6}}\right) \)
- \( \cos^{-1}\left(\frac{\sqrt{2}}{2}\right) \)
- \( \cos^{-1}\left(\frac{1}{3}\right) \)
- \( \cos^{-1}\left(\frac{\sqrt{2}}{3}\right) \)
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The Correct Option is D
Solution and Explanation
Concept:
\[
\cos\theta = \frac{\vec{d_1} \cdot \vec{d_2}}{|\vec{d_1}||\vec{d_2}|}
\]
Step 1: Direction vectors.
\[ \vec{d_1} = (1,-1,-1), \quad \vec{d_2} = (2,2,-2) \]
Step 2: Dot product.
\[ \vec{d_1} \cdot \vec{d_2} = 2 -2 +2 = 2 \]
Step 3: Magnitudes.
\[ |\vec{d_1}| = \sqrt{3}, \quad |\vec{d_2}| = 2\sqrt{3} \]
Step 4: Compute angle.
\[ \cos\theta = \frac{2}{\sqrt{3} \cdot 2\sqrt{3}} = \frac{2}{6} = \frac{1}{3} \]
Step 1: Direction vectors.
\[ \vec{d_1} = (1,-1,-1), \quad \vec{d_2} = (2,2,-2) \]
Step 2: Dot product.
\[ \vec{d_1} \cdot \vec{d_2} = 2 -2 +2 = 2 \]
Step 3: Magnitudes.
\[ |\vec{d_1}| = \sqrt{3}, \quad |\vec{d_2}| = 2\sqrt{3} \]
Step 4: Compute angle.
\[ \cos\theta = \frac{2}{\sqrt{3} \cdot 2\sqrt{3}} = \frac{2}{6} = \frac{1}{3} \]
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