Question:
If the lines \(\frac{2x-1}{2} = \frac{3-y}{1} = \frac{z-1}{3}\) and \(\frac{x+3}{2} = \frac{y+2}{5} = \frac{z+1}{a}\) are perpendicular to each other, then the value of \(a\) is:
If the lines \(\frac{2x-1}{2} = \frac{3-y}{1} = \frac{z-1}{3}\) and \(\frac{x+3}{2} = \frac{y+2}{5} = \frac{z+1}{a}\) are perpendicular to each other, then the value of \(a\) is:
Show Hint
Always extract direction ratios correctly before applying dot product condition.
Updated On: Apr 30, 2026
- \(\frac{1}{4} \)
- \(\frac{3}{4} \)
- \(\frac{1}{2} \)
- \(2 \)
- \(1 \)
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The Correct Option is
Solution and Explanation
Concept:
Two lines are perpendicular if their direction vectors satisfy:
\[
\vec{d_1} \cdot \vec{d_2} = 0
\]
Step 1: Find direction vectors.
From first line: \[ \vec{d_1} = (2,-1,3) \] From second line: \[ \vec{d_2} = (2,5,a) \]
Step 2: Apply perpendicular condition.
\[ (2)(2) + (-1)(5) + (3)(a) = 0 \] \[ 4 - 5 + 3a = 0 \Rightarrow -1 + 3a = 0 \Rightarrow a = 1 \]
Step 1: Find direction vectors.
From first line: \[ \vec{d_1} = (2,-1,3) \] From second line: \[ \vec{d_2} = (2,5,a) \]
Step 2: Apply perpendicular condition.
\[ (2)(2) + (-1)(5) + (3)(a) = 0 \] \[ 4 - 5 + 3a = 0 \Rightarrow -1 + 3a = 0 \Rightarrow a = 1 \]
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