Question:
If for some \(\alpha \in \mathbb{R}\), the lines
\(L_1 : \dfrac{x+1}{2} = \dfrac{y-2}{-1} = \dfrac{z-1}{1}\)
and
\(L_2 : \dfrac{x+2}{\alpha} = \dfrac{y+1}{5-\alpha} = \dfrac{z+1}{1}\)
are coplanar, then the line \(L_2\) passes through the point :
If for some \(\alpha \in \mathbb{R}\), the lines
\(L_1 : \dfrac{x+1}{2} = \dfrac{y-2}{-1} = \dfrac{z-1}{1}\)
and
\(L_2 : \dfrac{x+2}{\alpha} = \dfrac{y+1}{5-\alpha} = \dfrac{z+1}{1}\)
are coplanar, then the line \(L_2\) passes through the point :
\(L_1 : \dfrac{x+1}{2} = \dfrac{y-2}{-1} = \dfrac{z-1}{1}\)
and
\(L_2 : \dfrac{x+2}{\alpha} = \dfrac{y+1}{5-\alpha} = \dfrac{z+1}{1}\)
are coplanar, then the line \(L_2\) passes through the point :
Show Hint
Set the scalar triple product of the joining vector and the two direction vectors equal to zero to find \(\alpha\), then test the points.
Updated On: Nov 5, 2026
- \((10, 2, 2)\)
- \((2, -10, -2)\)
- \((10, -2, -2)\)
- \((-2, 10, 2)\)
Show Solution
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The Correct Option is B
Solution and Explanation
\(L_1\) passes through \((-1, 2, 1)\) with direction \((2, -1, 1)\); \(L_2\) passes through \((-2, -1, -1)\) with direction \((\alpha, 5-\alpha, 1)\).
Joining vector: \((-2+1,\ -1-2,\ -1-1) = (-1, -3, -2)\).
Coplanarity: \(\begin{vmatrix} -1 & -3 & -2 \\ 2 & -1 & 1 \\ \alpha & 5-\alpha & 1 \end{vmatrix} = 0\).
Expanding: \(-1[-1 - (5-\alpha)] + 3[2 - \alpha] - 2[2(5-\alpha) + \alpha] = (6 - \alpha) + 6 - 3\alpha - 20 + 2\alpha = -2\alpha - 8 = 0 \Rightarrow \alpha = -4\).
So \(L_2 : \dfrac{x+2}{-4} = \dfrac{y+1}{9} = \dfrac{z+1}{1}\).
Test \((2, -10, -2)\): \(\dfrac{4}{-4} = -1,\ \dfrac{-9}{9} = -1,\ \dfrac{-1}{1} = -1\). All equal, so it lies on \(L_2\).
Test \((10, 2, 2)\): \(\dfrac{12}{-4} = -3\) but \(\dfrac{3}{9} = \dfrac13\). No. \((10,-2,-2)\): \(-3,\ -\tfrac19,\ -1\). No. \((-2, 10, 2)\): \(0,\ \tfrac{11}{9},\ 3\). No.
Answer: (2) \((2, -10, -2)\)
Joining vector: \((-2+1,\ -1-2,\ -1-1) = (-1, -3, -2)\).
Coplanarity: \(\begin{vmatrix} -1 & -3 & -2 \\ 2 & -1 & 1 \\ \alpha & 5-\alpha & 1 \end{vmatrix} = 0\).
Expanding: \(-1[-1 - (5-\alpha)] + 3[2 - \alpha] - 2[2(5-\alpha) + \alpha] = (6 - \alpha) + 6 - 3\alpha - 20 + 2\alpha = -2\alpha - 8 = 0 \Rightarrow \alpha = -4\).
So \(L_2 : \dfrac{x+2}{-4} = \dfrac{y+1}{9} = \dfrac{z+1}{1}\).
Test \((2, -10, -2)\): \(\dfrac{4}{-4} = -1,\ \dfrac{-9}{9} = -1,\ \dfrac{-1}{1} = -1\). All equal, so it lies on \(L_2\).
Test \((10, 2, 2)\): \(\dfrac{12}{-4} = -3\) but \(\dfrac{3}{9} = \dfrac13\). No. \((10,-2,-2)\): \(-3,\ -\tfrac19,\ -1\). No. \((-2, 10, 2)\): \(0,\ \tfrac{11}{9},\ 3\). No.
Answer: (2) \((2, -10, -2)\)
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