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If $\vec{a}$ and $\vec{b}$ are two unit vectors, then the vector $(\vec{a}+\vec{b}) \times(\vec{a} \times \vec{b})$ is parallel to the vector
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The correct answer is:
$\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}$
We have $(\vec{a}+\vec{b}) \times(\vec{a} \times \vec{b})$
$\begin{array}{l}
=\vec{a} \times(\vec{a} \times \vec{b})+\vec{b} \times(\vec{a} \times \vec{b}) \\
=(\vec{a} \cdot \vec{b}) \vec{a}-(\vec{a} \cdot \vec{a}) \vec{b}+(\vec{b} \cdot \vec{b}) \vec{a}-(\vec{b} \cdot \vec{a}) \vec{b} \\
=(\vec{a} \cdot \vec{b})(\vec{a}-\vec{b}) \\
\quad+\vec{a}-\vec{b}\left(\vec{b} \cdot \vec{b}=\vec{b}^{2}=1, \vec{a} \cdot \vec{a}=\vec{a}^{2}=1\right) \\
=(\vec{a} \cdot \vec{b}+1)(\vec{a}-\vec{b}) \\
=x(\vec{a}-\vec{b}) \text { where } x=\vec{a} \cdot \vec{b}+1 \text { is a scalar }
\end{array}$
$\therefore$ The given vector is parallel to $\vec{a}-\vec{b}$.
$\begin{array}{l}
=\vec{a} \times(\vec{a} \times \vec{b})+\vec{b} \times(\vec{a} \times \vec{b}) \\
=(\vec{a} \cdot \vec{b}) \vec{a}-(\vec{a} \cdot \vec{a}) \vec{b}+(\vec{b} \cdot \vec{b}) \vec{a}-(\vec{b} \cdot \vec{a}) \vec{b} \\
=(\vec{a} \cdot \vec{b})(\vec{a}-\vec{b}) \\
\quad+\vec{a}-\vec{b}\left(\vec{b} \cdot \vec{b}=\vec{b}^{2}=1, \vec{a} \cdot \vec{a}=\vec{a}^{2}=1\right) \\
=(\vec{a} \cdot \vec{b}+1)(\vec{a}-\vec{b}) \\
=x(\vec{a}-\vec{b}) \text { where } x=\vec{a} \cdot \vec{b}+1 \text { is a scalar }
\end{array}$
$\therefore$ The given vector is parallel to $\vec{a}-\vec{b}$.
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