CBSE 2024 · Region 1 · Set 2 · Q34 · 5 marks
Equations of sides of a parallelogram ABCD are as follows : $\displaystyle \mathrm{AB}: \frac{\mathrm{x}+1}{1}=\frac{\mathrm{y}-2}{-2}=\frac{\mathrm{z}-1}{2}$ $\displaystyle \mathrm{BC}: \frac{\mathrm{x}-1}{3}=\frac{\mathrm{y}+2}{-5}=\frac{\mathrm{z}-5}{3}$ $\displaystyle \mathrm{CD}: \frac{\mathrm{x}-4}{1}=\frac{\mathrm{y}+7}{-2}=\frac{\mathrm{z}-8}{2}$ $\displaystyle \mathrm{DA}: \frac{\mathrm{x}-2}{3}=\frac{\mathrm{y}+3}{-5}=\frac{\mathrm{z}-4}{3}$ Find the equation of diagonal BD .
Marking-scheme solution
Let coordinates of B (from AB ) $\displaystyle =(\lambda-1,2-2 \lambda, 2 \lambda+1)$
Also coordinates of B (from BC$\displaystyle )=(3 \mu+1,-5 \mu-2,3 \mu+5)$
Three Dimensional GeometryEquation of a Line in SpaceApplylong_answerhard
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CBSE Class 12 Mathematics past-paper question from the 2024board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.