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Mathematics · 2023 · 2 marks
CBSE 2023 · Region 4 · Set 1 · Q25
(a) Find the vector equation of the line passing through the point (2,1,3) and perpendicular to both the lines (x -1)/1=(y -2)/2=(z -3)/3;…
(a)
Find the vector equation of the line passing through the point $\displaystyle (2,1,3)$ and perpendicular to both the lines \[\frac{\mathrm{x}-1}{1}=\frac{\mathrm{y}-2}{2}=\frac{\mathrm{z}-3}{3} ; \quad \frac{\mathrm{x}}{-3}=\frac{\mathrm{y}}{2}=\frac{\mathrm{z}}{5} . \]
OR
(b)
The equations of a line are $\displaystyle 5 \mathrm{x}-3=15 \mathrm{y}+7=3-10 \mathrm{z}$. Write the direction cosines of the line and find the coordinates of a point through which it passes.
Marking-scheme solution
(a)
Vector equation of the line passing through ( $\displaystyle 2,1,3$ ) is
$$\overrightarrow{\mathrm{r}}=($\displaystyle 2$ \hat{\imath}+\hat{\jmath}+$\displaystyle 3$ \hat{k})+\lambda(\mathrm{a} \hat{\imath}+\mathrm{b} \hat{\jmath}+\mathrm{c} \hat{k})
$$Line $\displaystyle \overrightarrow{\mathrm{r}}$ is perpendicular to the given lines then
$$\begin{aligned}
\mathrm{a}+$\displaystyle 2$ \mathrm{b} & +$\displaystyle 3$ \mathrm{c}=$\displaystyle 0$ ;-$\displaystyle 3$ \mathrm{a}+$\displaystyle 2$ \mathrm{b}+$\displaystyle 5$ \mathrm{c}=$\displaystyle 0$
\Rightarrow & \frac{\mathrm{a}}{$\displaystyle 4$}=\frac{\mathrm{b}}{-$\displaystyle 14$}=\frac{\mathrm{c}}{$\displaystyle 8$}=k^{\prime}(\text { say })
\Rightarrow \mathrm{a} & =$\displaystyle 2$ k, \mathrm{b}=-$\displaystyle 7$ k \text { and } \mathrm{c}=$\displaystyle 4$ k
\end{aligned}
$$Thus, the required vector equation is
$$\overrightarrow{\mathrm{r}}=($\displaystyle 2$ \hat{\imath}+\hat{\jmath}+$\displaystyle 3$ \hat{k})+\lambda($\displaystyle 2$ \hat{\imath}-$\displaystyle 7$ \hat{\jmath}+$\displaystyle 4$ \hat{k})
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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.