CBSE 2026 · Region 5 · Set 1 · Q35 · 5 marks
Find the foot of the perpendicular from the point $\displaystyle (0,2,3)$ on the line $\displaystyle \frac{-\mathrm{x}-3}{-5}=\frac{1-\mathrm{y}}{-2}=\frac{3 \mathrm{z}+12}{9}$ and hence find the length of the perpendicular.Find the value of p if the shortest distance between the lines \[\begin{aligned} & \vec{r}=(\hat{i}+2 \hat{j}+\hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k}) \text { and } \\ & \vec{r}=(p \hat{i}-\hat{j}-\hat{k})+\mu(2 \hat{i}+\hat{j}+2 \hat{k}) \\ & \text { is } \frac{3}{\sqrt{2}} \text { units. } \end{aligned} \]
Find the foot of the perpendicular from the point $\displaystyle (0,2,3)$ on the line $\displaystyle \frac{-\mathrm{x}-3}{-5}=\frac{1-\mathrm{y}}{-2}=\frac{3 \mathrm{z}+12}{9}$ and hence find the length of the perpendicular.
Find the value of p if the shortest distance between the lines \[\begin{aligned} & \vec{r}=(\hat{i}+2 \hat{j}+\hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k}) \text { and } \\ & \vec{r}=(p \hat{i}-\hat{j}-\hat{k})+\mu(2 \hat{i}+\hat{j}+2 \hat{k}) \\ & \text { is } \frac{3}{\sqrt{2}} \text { units. } \end{aligned} \]
Marking-scheme solution
Given line is $\displaystyle \dfrac{x+3}{5}=\dfrac{y-1}{2}=\dfrac{z+4}{3}=\lambda$
Any general point on line $\displaystyle l$ is $\displaystyle (5 \lambda-3,2 \lambda+1,3 \lambda-4)$
Drs of given line are $\displaystyle \langle 5,2,3\rangle$
Drs of perpendicular line are $\displaystyle \langle 5 \lambda-3,2 \lambda-1,3 \lambda-7\rangle$
As lines are perpendicular
$\displaystyle \therefore 5(5 \lambda-3)+2(2 \lambda-1)+3(3 \lambda-7)=0$ gives $\displaystyle \lambda=1$
$\displaystyle \therefore$ coordinates of foot of perpendicular are $\displaystyle (2,3,-1)$
length of perpendicular $\displaystyle =\sqrt{21}$
Let $\displaystyle \vec{a_{1}}=\hat{i}+2 \hat{j}+\hat{k}, \vec{a_{2}}=p \hat{i}-\hat{j}-\hat{k}$, and $\displaystyle \vec{b_{1}}=\hat{i}-\hat{j}+\hat{k}, \vec{b_{2}}=2 \hat{i}+\hat{j}+2 \hat{k}$
Here $\displaystyle \left(\vec{a_{2}}-\vec{a_{1}}\right)=(p-1) \hat{i}-3 \hat{j}-2 \hat{k}$
$\displaystyle \vec{b_{1}} \times \vec{b_{2}}=-3 \hat{i}+3 \hat{k}$
S.D. $\displaystyle =\left|\dfrac{\left(\vec{a_{2}}-\vec{a_{1}}\right) \cdot\left(\vec{b_{1}} \times \vec{b_{2}}\right)}{\left|\vec{b_{1}} \times \vec{b_{2}}\right|}\right|$
$\displaystyle \dfrac{-3 p-3}{3 \sqrt{2}}= \pm \dfrac{3}{\sqrt{2}}$
gives $\displaystyle p=-4,2$
Three Dimensional GeometryShortest Distance between Two LinesApplylong_answerhard
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CBSE Class 12 Mathematics past-paper question from the 2026board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.