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The vector $(\hat{i} \times \vec{a} \cdot \vec{b}) \hat{i}+(\hat{j} \times \vec{a} \vec{b}) \hat{j}+(\hat{k} \times \vec{a} \cdot \vec{b}) \hat{k}$ is equal to:
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Verified Answer
The correct answer is:
$\vec{a} \times \vec{b}$
$\vec{a} \times \vec{b}$
$(\hat{i} \times \vec{a} \cdot \vec{b}) \hat{i}+(\hat{j} \times \vec{a} \cdot \vec{b}) \hat{j}+(\hat{k} \times \vec{a} \cdot \vec{b}) \hat{k}$
$=(\hat{i} \cdot \vec{a} \times \vec{b}) \hat{i}+(\hat{j} \cdot \vec{a} \times \vec{b}) \hat{j}+(\hat{k} \cdot \vec{a} \times \vec{b}) \hat{k}$
$(\because \vec{a} \times \vec{b} \cdot \vec{c}=\vec{a} \cdot \vec{b} \times \vec{c})$
$=(\vec{a} \times \vec{b}) \hat{i}+(\vec{a} \times \vec{b}) \hat{j}+(\vec{a} \times \vec{b}) \hat{k}$
$==\vec{a} \times \vec{b}$
$=(\hat{i} \cdot \vec{a} \times \vec{b}) \hat{i}+(\hat{j} \cdot \vec{a} \times \vec{b}) \hat{j}+(\hat{k} \cdot \vec{a} \times \vec{b}) \hat{k}$
$(\because \vec{a} \times \vec{b} \cdot \vec{c}=\vec{a} \cdot \vec{b} \times \vec{c})$
$=(\vec{a} \times \vec{b}) \hat{i}+(\vec{a} \times \vec{b}) \hat{j}+(\vec{a} \times \vec{b}) \hat{k}$
$==\vec{a} \times \vec{b}$
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