CBSE Class 12 Mathematics 2024: question 34(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. $L_1$ passes through $(2,-1,1)$ and is parallel to $\hat{i}+\hat{j}+3\hat{k}$:

   $L_1:\vec{r}=(2\hat{i}-\hat{j}+\hat{k})+\lambda (\hat{i}+\hat{j}+3\hat{k})$.

   1 mark

2. Rearrange $L_2$ by collecting the terms in $\mu$:

   $L_2:\vec{r}=(\hat{i}+\hat{j}-2\hat{k})+\mu (2\hat{j}-\hat{k})$.

   ½ mark

3. Identify the vectors:

   ${\vec{a}}_1=2\hat{i}-\hat{j}+\hat{k}$, ${\vec{b}}_1=\hat{i}+\hat{j}+3\hat{k}$

   ${\vec{a}}_2=\hat{i}+\hat{j}-2\hat{k}$, ${\vec{b}}_2=2\hat{j}-\hat{k}$

   ½ mark

4. Compute the two vectors needed:

   ${\vec{a}}_2-{\vec{a}}_1=-\hat{i}+2\hat{j}-3\hat{k}$

   ${\vec{b}}_1\times {\vec{b}}_2=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 1& 3\\ 0& 2& -1\end{array}\right|=(-1-6)\hat{i}-(-1-0)\hat{j}+(2-0)\hat{k}=-7\hat{i}+\hat{j}+2\hat{k}$

   1½ marks

5. Find the shortest distance:

   $({\vec{a}}_2-{\vec{a}}_1)\cdot ({\vec{b}}_1\times {\vec{b}}_2)=7+2-6=3$

   $|{\vec{b}}_1\times {\vec{b}}_2|=\sqrt{49+1+4}=\sqrt{54}=3\sqrt{6}$

   $SD=\frac{|({\vec{a}}_2-{\vec{a}}_1)\cdot ({\vec{b}}_1\times {\vec{b}}_2)|}{|{\vec{b}}_1\times {\vec{b}}_2|}=\frac{3}{3\sqrt{6}}=\frac{1}{\sqrt{6}}$ units.

   1½ marks

Answer: Shortest distance $=\frac{1}{\sqrt{6}}$ units

Revise the chapter

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