Question:
The direction ratios of the line $x - y + z - 5 = 0 = x - 3y - 6$ are
The direction ratios of the line $x - y + z - 5 = 0 = x - 3y - 6$ are
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Direction of intersection line = $\vec{n_1} \times \vec{n_2}$.
Updated On: Jul 3, 2026
- 2, -4, 1
- $\frac{2}{\sqrt{21}}, \frac{-4}{\sqrt{21}}, \frac{1}{\sqrt{21}}$
- 3, 1, 2
- $\frac{3}{\sqrt{14}}, \frac{1}{\sqrt{14}}, \frac{-2}{\sqrt{14}}$
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The Correct Option is C
Solution and Explanation
Step 1: Concept
A line given as the intersection of two planes has a direction vector equal to the cross product of the normals of the two planes.
Step 2: Meaning
$\vec{n_1} = (1, -1, 1)$. $\vec{n_2} = (1, -3, 0)$.
Step 3: Analysis
$\vec{n_1} \times \vec{n_2} = \begin{vmatrix} i & j & k \\ 1 & -1 & 1 \\ 1 & -3 & 0 \end{vmatrix} = i(0 - (-3)) - j(0 - 1) + k(-3 - (-1))$. This equals $3i + j - 2k$.
Step 4: Conclusion
The direction ratios are $(3, 1, -2)$. Final Answer: (C)
A line given as the intersection of two planes has a direction vector equal to the cross product of the normals of the two planes.
Step 2: Meaning
$\vec{n_1} = (1, -1, 1)$. $\vec{n_2} = (1, -3, 0)$.
Step 3: Analysis
$\vec{n_1} \times \vec{n_2} = \begin{vmatrix} i & j & k \\ 1 & -1 & 1 \\ 1 & -3 & 0 \end{vmatrix} = i(0 - (-3)) - j(0 - 1) + k(-3 - (-1))$. This equals $3i + j - 2k$.
Step 4: Conclusion
The direction ratios are $(3, 1, -2)$. Final Answer: (C)
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