CBSE Class 12 Mathematics 2024: question 30

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Step-by-step solution

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1. Let $\vec{a}=2\hat{i}-\hat{j}+\hat{k}$, $\vec{b}=\hat{i}+\hat{j}-\hat{k}$. $\vec{c}=\vec{a}\times \vec{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& -1& 1\\ 1& 1& -1\end{array}\right|=\hat{i}(1-1)-\hat{j}(-2-1)+\hat{k}(2+1)=3\hat{j}+3\hat{k}$.

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2. $|\vec{c}|=3\sqrt{2}$, so $\hat{c}=\frac{\hat{j}+\hat{k}}{\sqrt{2}}$. Required vector $\vec{d}=4\hat{c}=2\sqrt{2}\hat{j}+2\sqrt{2}\hat{k}$ (or $-2\sqrt{2}\hat{j}-2\sqrt{2}\hat{k}$).

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3. Verification: $|\vec{d}|=\sqrt{8+8}=4$; $\vec{d}\cdot \vec{a}=0-2\sqrt{2}+2\sqrt{2}=0$; $\vec{d}\cdot \vec{b}=0+2\sqrt{2}-2\sqrt{2}=0$. Hence $\vec{d}\perp \vec{a}$ and $\vec{d}\perp \vec{b}$.

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Answer: $\pm (2\sqrt{2}\hat{j}+2\sqrt{2}\hat{k})$

Revise the chapter

• Chapter: Vector Algebra (notes, formulas, where marks are lost and every past question)