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(B) Given $a = x \hat{i} + y \hat{j} + z \hat{k}$.We need to evaluate the expression $E = (a 	imes \hat{i}) \cdot (\hat{i} + \hat{j}) + (a 	imes \ha

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$x+y+z$

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(B) Given $a = x \hat{i} + y \hat{j} + z \hat{k}$.We need to evaluate the expression $E = (a 	imes \hat{i}) \cdot (\hat{i} + \hat{j}) + (a 	imes \hat{j}) \cdot (\hat{j} + \hat{k}) + (a 	imes \hat{k}) \cdot (\hat{k} + \hat{i})$.Using the property of scalar triple product $[a, b, c] = (a 	imes b) \cdot c$, we can rewrite the expression as:$E = [a, \hat{i}, \hat{i}] + [a, \hat{i}, \hat{j}] + [a, \hat{j}, \hat{j}] + [a, \hat{j}, \hat{k}] + [a, \hat{k}, \hat{k}] + [a, \hat{k}, \hat{i}]$.Since the scalar triple product is zero if any two vectors are identical, we have $[a, \hat{i}, \hat{i}] = [a, \hat{j}, \hat{j}] = [a, \hat{k}, \hat{k}] = 0$.Thus, $E = [a, \hat{i}, \hat{j}] + [a, \hat{j}, \hat{k}] + [a, \hat{k}, \hat{i}]$.Using the cyclic property $[a, b, c] = a \cdot (b 	imes c)$, we get:$E = a \cdot (\hat{i} 	imes \hat{j}) + a \cdot (\hat{j} 	imes \hat{k}) + a \cdot (\hat{k} 	imes \hat{i})$.Since $\hat{i} 	imes \hat{j} = \hat{k}$, $\hat{j} 	imes \hat{k} = \hat{i}$, and $\hat{k} 	imes \hat{i} = \hat{j}$, we have:$E = a \cdot \hat{k} + a \cdot \hat{i} + a \cdot \hat{j} = a \cdot (\hat{i} + \hat{j} + \hat{k})$.Substituting $a = x \hat{i} + y \hat{j} + z \hat{k}$, we get:$E = (x \hat{i} + y \hat{j} + z \hat{k}) \cdot (\hat{i} + \hat{j} + \hat{k}) = x + y + z$.

If $a=x \hat{i}+y \hat{j}+z \hat{k}$, then $(a \times \hat{i}) \cdot(\hat{i}+\hat{j})+(a \times \hat{j}) \cdot(\hat{j}+\hat{k})+(a \times \hat{k}) \cdot(\hat{k}+\hat{i})=$

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