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If $\mathbf{a}=x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}$, then $(\mathbf{a} \times \hat{\mathbf{i}}) \cdot(\hat{\mathbf{i}}+\hat{\mathbf{j}})+(\mathbf{a} \times \hat{\mathbf{j}}) \cdot(\hat{\mathbf{j}}+\hat{\mathbf{k}})+(\mathbf{a} \times \hat{\mathbf{k}}) \cdot(\hat{\mathbf{k}}+\hat{\mathbf{i}})=$
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Verified Answer
The correct answer is:
$x+y+z$
We have,
$\begin{aligned}
& \mathbf{a}=x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}} \\
& (\mathbf{a} \times \hat{\mathbf{i}})(\hat{\mathbf{i}}+\hat{\mathbf{j}})+(\hat{\mathbf{a}} \times \hat{\mathbf{j}})(\hat{\mathbf{j}}+\hat{\mathbf{k}})+(\mathbf{a} \times \hat{\mathbf{k}})(\hat{\mathbf{k}}+\hat{\mathbf{i}}) \\
& =[\mathbf{a} \hat{\mathbf{i}} \hat{\mathbf{i}}]+[\mathbf{a} \hat{\mathbf{i}} \hat{\mathbf{j}}]+[\mathbf{a} \hat{\mathbf{j}} \hat{\mathbf{j}}]+[\mathbf{a} \hat{\mathbf{j}} \hat{\mathbf{k}}]+[\mathbf{a} \hat{\mathbf{k}} \hat{\mathbf{k}}]+[\mathbf{a} \hat{\mathbf{k}} \hat{\mathbf{i}}] \\
& =[\mathbf{a} \hat{\mathbf{i}} \hat{\mathbf{j}}]+[\mathbf{a} \hat{\mathbf{j}} \hat{\mathbf{k}}]+[\mathbf{a} \hat{\mathbf{k}} \hat{\mathbf{i}}] \\
& \qquad[\because[\mathbf{a} \hat{\mathbf{i}} \hat{\mathbf{i}}]=[\mathbf{a} \hat{\mathbf{j}} \hat{\mathbf{j}}]=[\mathbf{a} \hat{\mathbf{k}} \hat{\mathbf{k}}]=0] \\
& =\mathbf{a} \cdot(\hat{\mathbf{i}} \times \hat{\mathbf{j}})+\mathbf{a} \cdot(\hat{\mathbf{j}} \times \hat{\mathbf{k}})+\mathbf{a} \cdot(\hat{\mathbf{k}} \times \hat{\mathbf{i}}) \\
& =\mathbf{a} \cdot \hat{\mathbf{k}}+\mathbf{a} \cdot \hat{\mathbf{i}}+\mathbf{a} \cdot \hat{\mathbf{j}}=\mathbf{a} \cdot(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}) \\
& =(x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}) \cdot(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})=x+y+z
\end{aligned}$
$\begin{aligned}
& \mathbf{a}=x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}} \\
& (\mathbf{a} \times \hat{\mathbf{i}})(\hat{\mathbf{i}}+\hat{\mathbf{j}})+(\hat{\mathbf{a}} \times \hat{\mathbf{j}})(\hat{\mathbf{j}}+\hat{\mathbf{k}})+(\mathbf{a} \times \hat{\mathbf{k}})(\hat{\mathbf{k}}+\hat{\mathbf{i}}) \\
& =[\mathbf{a} \hat{\mathbf{i}} \hat{\mathbf{i}}]+[\mathbf{a} \hat{\mathbf{i}} \hat{\mathbf{j}}]+[\mathbf{a} \hat{\mathbf{j}} \hat{\mathbf{j}}]+[\mathbf{a} \hat{\mathbf{j}} \hat{\mathbf{k}}]+[\mathbf{a} \hat{\mathbf{k}} \hat{\mathbf{k}}]+[\mathbf{a} \hat{\mathbf{k}} \hat{\mathbf{i}}] \\
& =[\mathbf{a} \hat{\mathbf{i}} \hat{\mathbf{j}}]+[\mathbf{a} \hat{\mathbf{j}} \hat{\mathbf{k}}]+[\mathbf{a} \hat{\mathbf{k}} \hat{\mathbf{i}}] \\
& \qquad[\because[\mathbf{a} \hat{\mathbf{i}} \hat{\mathbf{i}}]=[\mathbf{a} \hat{\mathbf{j}} \hat{\mathbf{j}}]=[\mathbf{a} \hat{\mathbf{k}} \hat{\mathbf{k}}]=0] \\
& =\mathbf{a} \cdot(\hat{\mathbf{i}} \times \hat{\mathbf{j}})+\mathbf{a} \cdot(\hat{\mathbf{j}} \times \hat{\mathbf{k}})+\mathbf{a} \cdot(\hat{\mathbf{k}} \times \hat{\mathbf{i}}) \\
& =\mathbf{a} \cdot \hat{\mathbf{k}}+\mathbf{a} \cdot \hat{\mathbf{i}}+\mathbf{a} \cdot \hat{\mathbf{j}}=\mathbf{a} \cdot(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}) \\
& =(x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}) \cdot(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})=x+y+z
\end{aligned}$
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