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If $\vec{a}$ and $\vec{b}$ are vectors such that $|\vec{a}+\vec{b}|=\sqrt{29}$ and $\vec{a} \times(2 \hat{i}+3 \hat{j}+4 \hat{k})=(2 \hat{i}+3 \hat{j}+4 \hat{k}) \times \vec{b}$, then a possible value of $(\vec{a}+\vec{b}) \cdot(-7 \hat{i}+2 \hat{j}+3 \hat{k})$ is
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Verified Answer
The correct answer is:
4
Given that $\vec{a} \times(2 \hat{i}+3 \hat{j}+4 \hat{k})=(2 \hat{i}+3 \hat{j}+4 \hat{k}) \times \vec{b}$ $\Rightarrow(\vec{a}+\vec{b}) \times(2 \hat{i}+3 \hat{j}+4 \hat{k})=\overrightarrow{0}$
But $\vec{a}+\vec{b} \neq 0$ and $2 \hat{i}+3 \hat{j}+4 \hat{k} \neq 0$
$\therefore(\vec{a}+\vec{b}) \|(2 \hat{i}+3 \hat{j}+4 \hat{k})$.
Let $\vec{a}+\vec{b}=\lambda(2 \hat{i}+3 \hat{j}+4 \hat{k})$
Also given that $|\vec{a}+\vec{b}|=\sqrt{29} \Rightarrow \lambda=\pm 1$
$\therefore \vec{a}+\vec{b}=\pm(2 \hat{i}+3 \hat{j}+4 \hat{k})$
So, $(\vec{a}+\vec{b}) \cdot(-7 \hat{i}+2 \hat{i}+3 \hat{k})=\pm 4$
But $\vec{a}+\vec{b} \neq 0$ and $2 \hat{i}+3 \hat{j}+4 \hat{k} \neq 0$
$\therefore(\vec{a}+\vec{b}) \|(2 \hat{i}+3 \hat{j}+4 \hat{k})$.
Let $\vec{a}+\vec{b}=\lambda(2 \hat{i}+3 \hat{j}+4 \hat{k})$
Also given that $|\vec{a}+\vec{b}|=\sqrt{29} \Rightarrow \lambda=\pm 1$
$\therefore \vec{a}+\vec{b}=\pm(2 \hat{i}+3 \hat{j}+4 \hat{k})$
So, $(\vec{a}+\vec{b}) \cdot(-7 \hat{i}+2 \hat{i}+3 \hat{k})=\pm 4$
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