If a is a unit vector, then |a times hat{i}|^ | vedclass

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(A) Let $a = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}$. Since $a$ is a unit vector, $|a|^2 = a_1^2 + a_2^2 + a_3^2 = 1$. Now, $a 	imes \hat{i} = (a_1

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$2$

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(A) Let $a = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}$. Since $a$ is a unit vector, $|a|^2 = a_1^2 + a_2^2 + a_3^2 = 1$. Now, $a 	imes \hat{i} = (a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}) 	imes \hat{i} = -a_2 \hat{k} + a_3 \hat{j}$. So, $|a 	imes \hat{i}|^2 = a_2^2 + a_3^2$. Similarly, $|a 	imes \hat{j}|^2 = a_1^2 + a_3^2$ and $|a 	imes \hat{k}|^2 = a_1^2 + a_2^2$. Adding these, we get $|a 	imes \hat{i}|^2 + |a 	imes \hat{j}|^2 + |a 	imes \hat{k}|^2 = (a_2^2 + a_3^2) + (a_1^2 + a_3^2) + (a_1^2 + a_2^2) = 2(a_1^2 + a_2^2 + a_3^2)$. Since $a_1^2 + a_2^2 + a_3^2 = 1$, the sum is $2(1) = 2$.

If $a$ is a unit vector, then $|a \times \hat{i}|^2+|a \times \hat{j}|^2+|a \times \hat{k}|^2=$

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