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Mathematics · 2026 · 3 marks
CBSE 2026 · Region 4 · Set 1 · Q30
(a) Find a point on the line (x -2)/3=(1- y)/2=(z -3)/2 at a distance of √ 2 units from the point (1, 2, 3).
(a)
Find a point on the line $\displaystyle \frac{\mathrm{x}-2}{3}=\frac{1-\mathrm{y}}{2}=\frac{\mathrm{z}-3}{2}$ at a distance of $\displaystyle \sqrt{2}$ units from the point ($\displaystyle 1$, $\displaystyle 2$, $\displaystyle 3$).
OR
(b)
Find the shortest distance between the lines \[\begin{aligned} & \vec{r}=(4+\lambda) \hat{i}+(2 \lambda-1) \hat{j}-3 \lambda \hat{k} \\ & \vec{r}=(1+2 \mu) \hat{i}+(2-5 \mu) \hat{k}+(4 \mu-1) \hat{j} \end{aligned} \]
Marking-scheme solution
Let $\displaystyle \dfrac{x-2}{3}=\dfrac{y-1}{-2}=\dfrac{z-3}{2}=k$
A general point P on the line be $\displaystyle (3 k+2,-2 k+1,2 k+3)$
It is given that $\displaystyle PQ=\sqrt{2}$ where $\displaystyle Q(1,2,3)$, so
$\displaystyle PQ=\sqrt{(3 k+1)^{2}+(-2 k-1)^{2}+(2 k)^{2}}=\sqrt{2} \Rightarrow 17 k^{2}+10 k=0$
$\displaystyle \therefore k=0$ or $\displaystyle \dfrac{-10}{17}$
Thus, required points are $\displaystyle (2,1,3)$ & $\displaystyle \left(\dfrac{4}{17}, \dfrac{37}{17}, \dfrac{31}{17}\right)$
Here, $\displaystyle \vec{a_{1}}=4 \hat{i}-\hat{j}, \vec{b_{1}}=\hat{i}+2 \hat{j}-3 \hat{k}, \vec{a_{2}}=\hat{i}-\hat{j}+2 \hat{k}, \vec{b_{2}}=2 \hat{i}+4 \hat{j}-5 \hat{k}$
Now, $\displaystyle \vec{a_{2}}-\vec{a_{1}}=-3 \hat{i}+2 \hat{k}, \quad \vec{b_{1}} \times \vec{b_{2}}=2 \hat{i}-\hat{j}$ and $\displaystyle \left|\vec{b_{1}} \times \vec{b_{2}}\right|=\sqrt{5}$
Required distance $\displaystyle =\left|\dfrac{(-3 \hat{i}+2 \hat{k}) \cdot(2 \hat{i}-\hat{j})}{\sqrt{5}}\right|=\dfrac{6}{\sqrt{5}}$ or $\displaystyle \dfrac{6 \sqrt{5}}{5}$
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