Question:
Two lines
\(L_1:\;x=5,\; \dfrac{y}{3-\alpha}=\dfrac{z}{-2}\)
\(L_2:\;x=\alpha,\; \dfrac{y}{1}=\dfrac{z}{2-\alpha}\)
are coplanar. Then \(\alpha\) can take value(s)
Two lines
\(L_1:\;x=5,\; \dfrac{y}{3-\alpha}=\dfrac{z}{-2}\)
\(L_2:\;x=\alpha,\; \dfrac{y}{1}=\dfrac{z}{2-\alpha}\)
are coplanar. Then \(\alpha\) can take value(s)
Show Hint
Use scalar triple product for coplanarity.
Updated On: Mar 23, 2026
- \(1,4,5\)
- \(1,2,5\)
- \(3,4,5\)
- \(2,4,5\)
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The Correct Option is B
Solution and Explanation
Step 1: Coplanarity condition: \[ (\vec r_2-\vec r_1)\cdot(\vec d_1\times\vec d_2)=0 \]
Step 2: Solving determinant gives \(\alpha=1,2,5\).
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