CBSE Class 12 Mathematics sample paper 2023-24: question 35(b)

Hints

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Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

1. The direction vectors $\hat{i}-\hat{j}+\hat{k}$ and $-2\hat{j}+\hat{k}$ are not parallel, so the lines are non-parallel. A point $P$ on line 1 is $P(\lambda ,-\lambda ,\lambda )$ and a point $Q$ on line 2 is $Q(1,-1-2\mu ,\mu )$.

   ½ mark

2. $\overrightarrow{PQ}=(1-\lambda )\hat{i}+(-1-2\mu +\lambda )\hat{j}+(\mu -\lambda )\hat{k}$. For the shortest distance $PQ$ must be perpendicular to both lines.

   ½ mark

3. $\overrightarrow{PQ}\cdot (\hat{i}-\hat{j}+\hat{k})=0$: $(1-\lambda )-(-1-2\mu +\lambda )+(\mu -\lambda )=0\Rightarrow 2+3\mu -3\lambda =0$.

   ½ mark

4. $\overrightarrow{PQ}\cdot (-2\hat{j}+\hat{k})=0$: $-2(-1-2\mu +\lambda )+(\mu -\lambda )=0\Rightarrow 2+5\mu -3\lambda =0$.

   ½ mark

5. Subtracting: $2\mu =0\Rightarrow \mu =0$, and then $3\lambda =2\Rightarrow \lambda =\frac{2}{3}$.

   1 mark

6. The aeroplanes should be at $P\left(\frac{2}{3},-\frac{2}{3},\frac{2}{3}\right)$, i.e. $\frac{2}{3}(\hat{i}-\hat{j}+\hat{k})$, and at $Q(1,-1,0)$, i.e. $\hat{i}-\hat{j}$.

   1 mark

7. $\overrightarrow{PQ}=\frac{1}{3}\hat{i}-\frac{1}{3}\hat{j}-\frac{2}{3}\hat{k}$, so $|\overrightarrow{PQ}|=\sqrt{\frac{1}{9}+\frac{1}{9}+\frac{4}{9}}=\sqrt{\frac{2}{3}}$ units. This is the shortest distance.

   1 mark

Answer: Points $\left(\frac{2}{3},-\frac{2}{3},\frac{2}{3}\right)$ and $(1,-1,0)$; shortest distance $\sqrt{2/3}$ units

Revise the chapter

• Chapter: Three Dimensional Geometry (notes, formulas, where marks are lost and every past question)
• Topic: Shortest Distance Between Two Lines