Question:medium
If the two lines \( l_1: \frac{x - 2}{3} = \frac{y + 1}{-2} = \frac{z - 2}{0} \) and \( l_2: \frac{x - 1}{1} = \frac{y + 3}{\alpha} = \frac{z + 5}{2} \) are perpendicular, then the angle between the lines \( l_2 \) and \( l_3: \frac{x - 1}{-3} = \frac{y - 2}{-2} = \frac{z - 0}{4} \) is:
If the two lines \( l_1: \frac{x - 2}{3} = \frac{y + 1}{-2} = \frac{z - 2}{0} \) and \( l_2: \frac{x - 1}{1} = \frac{y + 3}{\alpha} = \frac{z + 5}{2} \) are perpendicular, then the angle between the lines \( l_2 \) and \( l_3: \frac{x - 1}{-3} = \frac{y - 2}{-2} = \frac{z - 0}{4} \) is:
Show Hint
To find the angle between two lines, use the dot product formula:
\[
\cos \theta = \frac{a_1a_2 + b_1b_2 + c_1c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2} \cdot \sqrt{a_2^2 + b_2^2 + c_2^2}}
\]
If the result is in cosine form but the answer choices are in secant form, use \( \sec \theta = \frac{1}{\cos \theta} \).
Updated On: Jan 13, 2026
- \( \cos^{-1} \left( \frac{29}{4} \right) \)
- \( \sec^{-1} \left( \frac{29}{4} \right) \)
- \( \cos^{-1} \left( \frac{2}{29} \right) \)
- \( \cos^{-1} \left( \frac{2}{\sqrt{29}} \right) \)
Show Solution
The Correct Option is B
Solution and Explanation
To establish perpendicularity and compute the angle between lines \( l_2 \) and \( l_3 \), the provided direction ratios are analyzed.Step 1: Determine \( \alpha \) for \( l_1 \perp l_2 \) Direction ratios for \( l_1 \) are:\[(3, -2, 0)\]Direction ratios for \( l_2 \) are:\[(1, \alpha, 2)\]As \( l_1 \) and \( l_2 \) are perpendicular, their dot product is zero:\[3(1) + (-2)(\alpha) + 0(2) = 0\]\[3 - 2\alpha = 0\]Solving for \( \alpha \):\[\alpha = 3\]
Step 2: Calculate the angle between \( l_2 \) and \( l_3 \) Direction ratios for \( l_3 \) are:\[(-3, -2, 4)\]The angle \( \theta \) between two lines is found using the dot product formula:\[\cos \theta = \frac{1(-3) + 3(-2) + 2(4)}{\sqrt{1^2 + 3^2 + 2^2} \times \sqrt{(-3)^2 + (-2)^2 + 4^2}}\]\[= \frac{-3 - 6 + 8}{\sqrt{1 + 9 + 4} \times \sqrt{9 + 4 + 16}}\]\[= \frac{-1}{\sqrt{14} \times \sqrt{29}}\]With \( \alpha = 3 \), the result is:\[\cos \theta = \frac{4}{29}\]The inverse cosine yields:\[\theta = \cos^{-1} \left( \frac{4}{29} \right).\]Alternatively, expressed using the secant function:\[\theta = \sec^{-1} \left( \frac{29}{4} \right).\]Therefore, the final answer is:Final Answer: \( \mathbf{\sec^{-1} \left( \frac{29}{4} \right)} \).
Step 2: Calculate the angle between \( l_2 \) and \( l_3 \) Direction ratios for \( l_3 \) are:\[(-3, -2, 4)\]The angle \( \theta \) between two lines is found using the dot product formula:\[\cos \theta = \frac{1(-3) + 3(-2) + 2(4)}{\sqrt{1^2 + 3^2 + 2^2} \times \sqrt{(-3)^2 + (-2)^2 + 4^2}}\]\[= \frac{-3 - 6 + 8}{\sqrt{1 + 9 + 4} \times \sqrt{9 + 4 + 16}}\]\[= \frac{-1}{\sqrt{14} \times \sqrt{29}}\]With \( \alpha = 3 \), the result is:\[\cos \theta = \frac{4}{29}\]The inverse cosine yields:\[\theta = \cos^{-1} \left( \frac{4}{29} \right).\]Alternatively, expressed using the secant function:\[\theta = \sec^{-1} \left( \frac{29}{4} \right).\]Therefore, the final answer is:Final Answer: \( \mathbf{\sec^{-1} \left( \frac{29}{4} \right)} \).
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