Question:
Two lines:
L₁: \(x = 5, \; \frac{y}{3 - \alpha} = \frac{z}{-2}\)
L₂: \(x = \alpha, \; \frac{y}{-1} = \frac{z}{2 - \alpha}\)
are coplanar. Then \(\alpha\) can take value(s):
Two lines:
L₁: \(x = 5, \; \frac{y}{3 - \alpha} = \frac{z}{-2}\)
L₂: \(x = \alpha, \; \frac{y}{-1} = \frac{z}{2 - \alpha}\)
are coplanar. Then \(\alpha\) can take value(s):
Show Hint
Two lines are coplanar if the scalar triple product is zero.
Updated On: Mar 23, 2026
- \(1,4,5\)
- \(1,2,5\)
- \(3,4,5\)
- 2,4,5
Show Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Using the coplanarity condition of two lines and solving the determinant, we get
α=1,2,5
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