Question:
If the lines \( \frac{x+3}{-3} = \frac{y-1}{k} = \frac{z-5}{5} \) and \( \frac{x+1}{-1} = \frac{y-2}{2} = \frac{z-5}{5} \) are coplanar, then the value of \( k \) is
If the lines \( \frac{x+3}{-3} = \frac{y-1}{k} = \frac{z-5}{5} \) and \( \frac{x+1}{-1} = \frac{y-2}{2} = \frac{z-5}{5} \) are coplanar, then the value of \( k \) is
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Scalar triple product zero ⇒ coplanar lines.
Updated On: Apr 30, 2026
- \(1\)
- \(2\)
- \(3\)
- \(4\)
- \(5\)
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The Correct Option is B
Solution and Explanation
Concept:
Lines are coplanar if scalar triple product = 0.
Step 1: Extract direction vectors. \[ \vec{d_1}=(-3, k, 5), \vec{d_2}=(-1,2,5) \]
Step 2: Take points. \[ P(-3,1,5), Q(-1,2,5) \] \[ \vec{PQ} = (2,1,0) \]
Step 3: Apply coplanarity condition. \[ [\vec{d_1},\vec{d_2},\vec{PQ}] = 0 \] \[ \begin{vmatrix} -3 & k & 5 \\ -1 & 2 & 5 \\ 2 & 1 & 0 \end{vmatrix} = 0 \]
Step 4: Expand determinant. \[ = -3(2\cdot0 - 5\cdot1) - k(-1\cdot0 - 5\cdot2) + 5(-1\cdot1 - 2\cdot2) \] \[ = -3(-5) - k(-10) + 5(-1-4) \] \[ = 15 + 10k -25 \] \[ = 10k -10 = 0 \]
Step 5: Solve. \[ k=1 \]
Step 1: Extract direction vectors. \[ \vec{d_1}=(-3, k, 5), \vec{d_2}=(-1,2,5) \]
Step 2: Take points. \[ P(-3,1,5), Q(-1,2,5) \] \[ \vec{PQ} = (2,1,0) \]
Step 3: Apply coplanarity condition. \[ [\vec{d_1},\vec{d_2},\vec{PQ}] = 0 \] \[ \begin{vmatrix} -3 & k & 5 \\ -1 & 2 & 5 \\ 2 & 1 & 0 \end{vmatrix} = 0 \]
Step 4: Expand determinant. \[ = -3(2\cdot0 - 5\cdot1) - k(-1\cdot0 - 5\cdot2) + 5(-1\cdot1 - 2\cdot2) \] \[ = -3(-5) - k(-10) + 5(-1-4) \] \[ = 15 + 10k -25 \] \[ = 10k -10 = 0 \]
Step 5: Solve. \[ k=1 \]
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