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Mathematics · 2023 · 3 marks
CBSE 2023 · Region 5 · Set 3 · Q28
(a) Find the coordinates of the foot of the perpendicular drawn from point (5,7,3) to the line (x-15)/3=(y-29)/8=(z-5)/-5.
Find the coordinates of the foot of the perpendicular drawn from point $\displaystyle (5,7,3)$ to the line $\displaystyle \frac{x-15}{3}=\frac{y-29}{8}=\frac{z-5}{-5}$.
If $\displaystyle \overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}$ and $\displaystyle \overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$ then find a unit vector perpendicular to both $\displaystyle \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}$ and $\displaystyle \overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}$.
Marking-scheme solution
(a)
Line: $\displaystyle (15+3t,\,29+8t,\,5-5t)$, direction $\displaystyle (3,8,-5)$. The vector from $\displaystyle (5,7,3)$ to a general point is $\displaystyle (10+3t,\,22+8t,\,2-5t)$; setting its dot product with $\displaystyle (3,8,-5)$ to zero: $\displaystyle 196+98t=0\Rightarrow t=-2$. Foot of perpendicular $\displaystyle =(9,\,13,\,15)$.
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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.