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Question

Let $\overrightarrow{\mathrm{a}}=\mathrm{a}_1 \hat{i}+\mathrm{a}_2 \hat{j}+\mathrm{a}_3 \hat{k}$ and $\overrightarrow{\mathrm{b}}=\mathrm{b}_1 \hat{i}+\mathrm{b}_2 \hat{j}+\mathrm{b}_3 \hat{k}$ be two vectors such that $|\overrightarrow{\mathrm{a}}|=1, \vec{a} \cdot \vec{b}=2$ and $|\vec{b}|=4$. If $\vec{c}=2(\vec{a} \times \vec{b})-3 \vec{b}$, then the angle between $\vec{b}$ and $\vec{c}$ is equal to:

$\cos ^{-1}\left(-\frac{1}{\sqrt{3}}\right)$
$\cos ^{-1}\left(\frac{2}{3}\right)$
$\cos ^{-1}\left(\frac{2}{\sqrt{3}}\right)$
$\cos ^{-1}\left(-\frac{\sqrt{3}}{2}\right)$

Solution

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