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If $\mathbf{a}$ and $\mathbf{b}$ are unit vectors, then the vector $(\mathbf{a}+\mathbf{b}) \times(\mathbf{a} \times \mathbf{b})$ is parallel to the vector
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Verified Answer
The correct answer is:
$\mathbf{a}+\mathbf{b}$
Now, $(\mathbf{a}+\mathbf{b}) \times(\mathbf{a} \times \mathbf{b})$
$\begin{aligned}
& \Rightarrow \mathbf{a} \times(\mathbf{a} \times \mathbf{b})+\mathbf{b} \times(\mathbf{a} \times \mathbf{b}) \\
& =(\mathbf{a} \cdot \mathbf{b}) \mathbf{a}-(\mathbf{a} \cdot \mathbf{a}) \mathbf{b}+(\mathbf{b} \cdot \mathbf{b}) \mathbf{a}-(\mathbf{b} \cdot \mathbf{a}) \mathbf{b} \\
& {[\because \mathbf{a} \text { and } \mathbf{b} \text { are unit vectors } \therefore \mathbf{a} . \mathbf{a}=\mathbf{b} . \mathbf{b}=1]} \\
& =(\mathbf{a} \cdot \mathbf{b}) \mathbf{a}-\mathbf{b}+\mathbf{a}-(\mathbf{b} \cdot \mathbf{a}) \mathbf{b} \\
& =(\mathbf{a} \cdot \mathbf{b})(\mathbf{a}-\mathbf{b})+\mathbf{a}-\mathbf{b} \\
& =(\mathbf{a}-\mathbf{b})(\mathbf{a} \cdot \mathbf{b}-1) \\
&
\end{aligned}$
$\therefore$ Given vector is parallel to $(\mathbf{a}-\mathbf{b})$.
$\begin{aligned}
& \Rightarrow \mathbf{a} \times(\mathbf{a} \times \mathbf{b})+\mathbf{b} \times(\mathbf{a} \times \mathbf{b}) \\
& =(\mathbf{a} \cdot \mathbf{b}) \mathbf{a}-(\mathbf{a} \cdot \mathbf{a}) \mathbf{b}+(\mathbf{b} \cdot \mathbf{b}) \mathbf{a}-(\mathbf{b} \cdot \mathbf{a}) \mathbf{b} \\
& {[\because \mathbf{a} \text { and } \mathbf{b} \text { are unit vectors } \therefore \mathbf{a} . \mathbf{a}=\mathbf{b} . \mathbf{b}=1]} \\
& =(\mathbf{a} \cdot \mathbf{b}) \mathbf{a}-\mathbf{b}+\mathbf{a}-(\mathbf{b} \cdot \mathbf{a}) \mathbf{b} \\
& =(\mathbf{a} \cdot \mathbf{b})(\mathbf{a}-\mathbf{b})+\mathbf{a}-\mathbf{b} \\
& =(\mathbf{a}-\mathbf{b})(\mathbf{a} \cdot \mathbf{b}-1) \\
&
\end{aligned}$
$\therefore$ Given vector is parallel to $(\mathbf{a}-\mathbf{b})$.
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