CBSE 2025 · Region 6 · Set 2 · Q28 · 3 marks
Find the distance of the point $\displaystyle \mathrm{P}(2,4,-1)$ from the line $\displaystyle \frac{\mathrm{x}+5}{1}=\frac{\mathrm{y}+3}{4}=\frac{\mathrm{z}-6}{-9}$.Let the position vectors of the points $\displaystyle \mathrm{A}, \mathrm{B}$ and C be $\displaystyle 3 \hat{\mathrm{i}}-\hat{\mathrm{j}}-2 \hat{\mathrm{k}}$, $\displaystyle \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}}$ and $\displaystyle \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$ respectively. Find the vector and cartesian equations of the line passing through A and parallel to line BC.
Find the distance of the point $\displaystyle \mathrm{P}(2,4,-1)$ from the line $\displaystyle \frac{\mathrm{x}+5}{1}=\frac{\mathrm{y}+3}{4}=\frac{\mathrm{z}-6}{-9}$.
Let the position vectors of the points $\displaystyle \mathrm{A}, \mathrm{B}$ and C be $\displaystyle 3 \hat{\mathrm{i}}-\hat{\mathrm{j}}-2 \hat{\mathrm{k}}$, $\displaystyle \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}}$ and $\displaystyle \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$ respectively. Find the vector and cartesian equations of the line passing through A and parallel to line BC.
Marking-scheme solution
Let $\displaystyle \vec{a}_{2}=2 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}-\hat{\mathrm{k}}, \overrightarrow{a_{1}}=-5 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+6 \hat{\mathrm{k}}$ and $\displaystyle \vec{b}=\hat{\mathrm{i}}+4 \hat{\mathrm{j}}-9 \hat{\mathrm{k}}$
Distance between point and line is given by $\displaystyle \mathrm{d}=\frac{\left|\left(\overrightarrow{\boldsymbol{a}_{\mathbf{2}}}-\overrightarrow{\boldsymbol{a}_{\mathbf{1}}}\right) \times \overrightarrow{\boldsymbol{b}}\right|}{|\overrightarrow{\boldsymbol{b}}|}$
Here $\displaystyle \left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)=7 \hat{\mathrm{i}}+7 \hat{\mathrm{j}}-7 \hat{\mathrm{k}}$
\[\begin{aligned}
& \left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right) \times \vec{b}=-35 \hat{\mathrm{i}}+56 \hat{\mathrm{j}}+21 \hat{\mathrm{k}} \\
& \mathbf{d}=\frac{49 \sqrt{2}}{7 \sqrt{2}}=7
\end{aligned}
\]
Direction vector of line $\displaystyle =3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}$
Vector equation is $\displaystyle \overrightarrow{\boldsymbol{r}}=3 \hat{\mathrm{i}}-\hat{\mathrm{j}}-2 \hat{\mathrm{k}}+\boldsymbol{\mu}(\mathbf{3} \hat{\mathrm{j}}+\mathbf{4} \hat{\mathrm{k}})$
Cartesian equation is $\displaystyle \frac{\mathrm{x}-3}{0}=\frac{\mathrm{y}+1}{3}=\frac{\mathrm{z}+2}{4}$
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CBSE Class 12 Mathematics past-paper question from the 2025board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.