Question:
If the lines \(\frac{x-1}{2} = \frac{y+1}{3} = \frac{z-1}{4}\) and \(\frac{x-2}{1} = \frac{y+m}{2} = \frac{z-2}{1}\) intersect each other, then the value of \(m\) is....
If the lines \(\frac{x-1}{2} = \frac{y+1}{3} = \frac{z-1}{4}\) and \(\frac{x-2}{1} = \frac{y+m}{2} = \frac{z-2}{1}\) intersect each other, then the value of \(m\) is....
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Lines intersect when the scalar triple product of the join vector and both directions is zero.
Updated On: Oct 1, 2026
- \(1\)
- \(2\)
- \(-1\)
- \(4\)
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The Correct Option is C
Solution and Explanation
Step 1: Points and directions
Line 1 passes through \((1,-1,1)\) with direction \((2,3,4)\). Line 2 passes through \((2,-m,2)\) with direction \((1,2,1)\).
Join vector: \((1,1-m,1)\).
Step 2: Coplanarity determinant
\[ \begin{vmatrix}1&1-m&1\\2&3&4\\1&2&1\end{vmatrix}=0 \]
Step 3: Expand
\(1(3-8)-(1-m)(2-4)+1(4-3)=-5+2(1-m)+1=-2-2m\).
Setting it to \(0\) gives \(m=-1\). Option (C).
Final Answer:
\(m=-1\), option (C). \[ \boxed{\text{(C) }-1} \]
Line 1 passes through \((1,-1,1)\) with direction \((2,3,4)\). Line 2 passes through \((2,-m,2)\) with direction \((1,2,1)\).
Join vector: \((1,1-m,1)\).
Step 2: Coplanarity determinant
\[ \begin{vmatrix}1&1-m&1\\2&3&4\\1&2&1\end{vmatrix}=0 \]
Step 3: Expand
\(1(3-8)-(1-m)(2-4)+1(4-3)=-5+2(1-m)+1=-2-2m\).
Setting it to \(0\) gives \(m=-1\). Option (C).
Final Answer:
\(m=-1\), option (C). \[ \boxed{\text{(C) }-1} \]
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