CBSE 2025 · Region 6 · Set 2 · Q25 · 2 marks
Find the value of $\displaystyle \lambda$ if the following lines are perpendicular to each other : \[\begin{aligned} & \mathrm{l}_{1}: \frac{1-\mathrm{x}}{-3}=\frac{3 \mathrm{y}-2}{2 \lambda}=\frac{\mathrm{z}-3}{3} \\ & \mathrm{l}_{2}: \frac{\mathrm{x}-1}{3 \lambda}=\frac{1-\mathrm{y}}{1}=\frac{2 \mathrm{z}-5}{3} \end{aligned} \]
Marking-scheme solution
$\displaystyle \mathrm{l}_{1}: \frac{\mathrm{x}-1}{3}=\frac{\mathrm{y}-\dfrac{2}{3}}{\dfrac{2}{3} \lambda}=\frac{\mathrm{z}-3}{3}$
\[\mathrm{l}_{2}: \frac{\mathrm{x}-1}{3 \lambda}=\frac{\mathrm{y}-1}{-1}=\frac{\mathrm{z}-\dfrac{5}{2}}{\dfrac{3}{2}}
\]
lines are perpendicular $\displaystyle \Rightarrow 3(3 \lambda)+\frac{2}{3} \lambda(-1)+3 \cdot \frac{3}{2}=0$
\[\lambda=\frac{-27}{50}
\]
Three Dimensional GeometryAngle between Two LinesApplyvery_short_answermedium
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CBSE Class 12 Mathematics past-paper question from the 2025board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.