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Let \(\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}\). Then \(|\vec{a}|^2 = a_1^2 + a_2^2 + a_3^2 = a^2\).
Now, we compute the cross products:
\(\vec{a} \times \hat{i} = (a_1\hat{i} + a_2\hat{j} + a_3\hat{k}) \times \hat{i} = a_2(\hat{j} \times \hat{i}) + a_3(\hat{k} \times \hat{i}) = -a_2\hat{k} + a_3\hat{j}\)
\(|\vec{a} \times \hat{i}|^2 = a_2^2 + a_3^2\)
\(\vec{a} \times \hat{j} = (a_1\hat{i} + a_2\hat{j} + a_3\hat{k}) \times \hat{j} = a_1(\hat{i} \times \hat{j}) + a_3(\hat{k} \times \hat{j}) = a_1\hat{k} - a_3\hat{i}\)
\(|\vec{a} \times \hat{j}|^2 = a_1^2 + a_3^2\)
\(\vec{a} \times \hat{k} = (a_1\hat{i} + a_2\hat{j} + a_3\hat{k}) \times \hat{k} = a_1(\hat{i} \times \hat{k}) + a_2(\hat{j} \times \hat{k}) = -a_1\hat{j} + a_2\hat{i}\)
\(|\vec{a} \times \hat{k}|^2 = a_1^2 + a_2^2\)
Adding these squared magnitudes:
\(|\vec{a} \times \hat{i}|^2 + |\vec{a} \times \hat{j}|^2 + |\vec{a} \times \hat{k}|^2 = (a_2^2 + a_3^2) + (a_1^2 + a_3^2) + (a_1^2 + a_2^2) = 2(a_1^2 + a_2^2 + a_3^2) = 2a^2\)
Correct Answer: 2a^{2}
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