The value of hat{i} times (hat{i} times vec{a | vedclass

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(C) Using the vector triple product identity $\vec{A} 	imes (\vec{B} 	imes \vec{C}) = \vec{B}(\vec{A} \cdot \vec{C}) - \vec{C}(\vec{A} \cdot \vec{B}

Vedclass · Accepted Answer

$-2\vec{a}$

Vedclass · Answer

(C) Using the vector triple product identity $\vec{A} 	imes (\vec{B} 	imes \vec{C}) = \vec{B}(\vec{A} \cdot \vec{C}) - \vec{C}(\vec{A} \cdot \vec{B})$:$\hat{i} 	imes (\hat{i} 	imes \vec{a}) = \hat{i}(\hat{i} \cdot \vec{a}) - \vec{a}(\hat{i} \cdot \hat{i}) = \hat{i}a_x - \vec{a}$$\hat{j} 	imes (\hat{j} 	imes \vec{a}) = \hat{j}(\hat{j} \cdot \vec{a}) - \vec{a}(\hat{j} \cdot \hat{j}) = \hat{j}a_y - \vec{a}$$\hat{k} 	imes (\hat{k} 	imes \vec{a}) = \hat{k}(\hat{k} \cdot \vec{a}) - \vec{a}(\hat{k} \cdot \hat{k}) = \hat{k}a_z - \vec{a}$Summing these expressions:$(\hat{i}a_x + \hat{j}a_y + \hat{k}a_z) - 3\vec{a}$Since $\vec{a} = \hat{i}a_x + \hat{j}a_y + \hat{k}a_z$,the expression becomes:$\vec{a} - 3\vec{a} = -2\vec{a}$.

The value of $\hat{i} \times (\hat{i} \times \vec{a}) + \hat{j} \times (\hat{j} \times \vec{a}) + \hat{k} \times (\hat{k} \times \vec{a})$ is

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