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Mathematics · 2023 · 3 marks
CBSE 2023 · Region 5 · Set 1 · Q30
(a) Find the coordinates of the foot of the perpendicular drawn from the point P (0,2,3) to the line (x+3)/5=(y -1)/2=(z +4)/3.
Find the coordinates of the foot of the perpendicular drawn from the point $\displaystyle \mathrm{P}(0,2,3)$ to the line $\displaystyle \frac{x+3}{5}=\frac{\mathrm{y}-1}{2}=\frac{\mathrm{z}+4}{3}$.
Three vectors $\displaystyle \vec{\mathrm{a}}, \vec{\mathrm{b}}$ and $\displaystyle \vec{\mathrm{c}}$ satisfy the condition $\displaystyle \vec{\mathrm{a}}+\vec{\mathrm{b}}+\vec{\mathrm{c}}=\overrightarrow{0}$. Evaluate the quantity $\displaystyle \mu=\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}}$, if $\displaystyle |\overrightarrow{\mathrm{a}}|=3$, $\displaystyle |\overrightarrow{\mathrm{b}}|=4$ and $\displaystyle |\overrightarrow{\mathrm{c}}|=2$.
Marking-scheme solution
General point on the given line is $\displaystyle \mathrm{M}(5 \lambda-3,2 \lambda+1,3 \lambda-4)$
Direction ratios of PM are $\displaystyle 5 \lambda-3,2 \lambda-1,3 \lambda-7$
$$\begin{aligned}
& (\vec{\mathrm{a}}+\vec{\mathrm{b}}+\vec{\mathrm{c}})^{2}=0 \\
& \Rightarrow \vec{\mathrm{a}}^{2}+\vec{\mathrm{b}}^{2}+\vec{\mathrm{c}}^{2}+2(\mu)=0 \\
& \Rightarrow \mu=-\frac{\mathbf{2 9}}{\mathbf{2}}
\end{aligned}
$$
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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.