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Mathematics · 2023 · 2 marks
CBSE 2023 · Region 3 · Set 2 · Q25
Find the value of p, so that lines (x-1)/-2=(y-4)/(3 p)=(z-3)/4 and (x-2)/(4 p)=(y-5)/2=(1-z)/7 are perpendicular to each other.
Find the value of $\displaystyle \mathrm{p}$, so that lines $\displaystyle \frac{x-1}{-2}=\frac{y-4}{3 \mathrm{p}}=\frac{z-3}{4}$ and $\displaystyle \frac{x-2}{4 \mathrm{p}}=\frac{y-5}{2}=\frac{1-z}{7}$ are perpendicular to each other.
Marking-scheme solution
d.r.'s of lines are < - $\displaystyle 2$, 3p, $\displaystyle 4>$ and < 4p, $\displaystyle 2$, - $\displaystyle 7>$
As lines are perpendicular
$$-8 \mathrm{p}+6 \mathrm{p}-28=0
$$
$\displaystyle \Rightarrow \mathrm{p}=-14$
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CBSE Class 12 Mathematics past-paper question from the 2023 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.