Question:

Let \( L_1 \) be the line that passes through the point \( (1, 2, -4) \) and be perpendicular to the two lines \( \frac{x - 8}{3} = \frac{y + 19}{-16} = \frac{z - 10}{7} \) and \( \frac{x - 15}{3} = \frac{y - 29}{8} = \frac{z - 5}{-5} \). If \( L_2 \) is the line \( \vec{r} = 7\hat{i} - 6\hat{k} + \lambda(\hat{i} + 2\hat{j} + 2\hat{k}) \), \( \lambda \in \mathbf{R} \), then an angle between \( L_1 \) and \( L_2 \) is equal to :

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The direction of L1 is the cross product of the two given direction vectors; then use the dot product with the direction of L2.
Updated On: Nov 12, 2026
  • \( \frac{\pi}{6} \)
  • \( \frac{\pi}{12} \)
  • \( \cos^{-1}\left(\frac{4}{5}\right) \)
  • \( \cos^{-1}\left(\frac{20}{21}\right) \)
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The Correct Option is D

Solution and Explanation

Direction vectors of the two given lines: \( (3, -16, 7) \) and \( (3, 8, -5) \).
\( \vec{b}_1 = (3, -16, 7) \times (3, 8, -5) = (80 - 56, -(-15 - 21), 24 + 48) = (24, 36, 72) \), which is proportional to \( (2, 3, 6) \).
Direction of L2: \( (1, 2, 2) \).
\( \cos\theta = \frac{2 + 6 + 12}{\sqrt{4 + 9 + 36}\,\sqrt{1 + 4 + 4}} = \frac{20}{7 \times 3} = \frac{20}{21} \).
So \( \theta = \cos^{-1}\left(\frac{20}{21}\right) \).
Answer: (4) \( \cos^{-1}\left(\frac{20}{21}\right) \)
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