The value of hat{k} cdot (hat{i} times hat{j} | vedclass

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(D) We know the properties of unit vectors in a Cartesian coordinate system: $\hat{i} 	imes \hat{j} = \hat{k}$ $\hat{j} 	imes \hat{k} = \hat{i}$ Sub

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$0$

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(D) We know the properties of unit vectors in a Cartesian coordinate system: $\hat{i} 	imes \hat{j} = \hat{k}$ $\hat{j} 	imes \hat{k} = \hat{i}$ Substituting these into the expression: $\hat{k} \cdot (\hat{i} 	imes \hat{j}) + \hat{i} \cdot (\hat{j} 	imes \hat{k}) = \hat{k} \cdot \hat{k} + \hat{i} \cdot \hat{i}$ Since the dot product of a unit vector with itself is $1$: $\hat{k} \cdot \hat{k} = 1$ $\hat{i} \cdot \hat{i} = 1$ Therefore,$1 + 1 = 2$. Wait,re-evaluating the expression: $\hat{k} \cdot (\hat{i} 	imes \hat{j}) + \hat{i} \cdot (\hat{j} 	imes \hat{k}) = \hat{k} \cdot \hat{k} + \hat{i} \cdot \hat{i} = 1 + 1 = 2$. Given the options,if the expression was $\hat{k} \cdot (\hat{i} 	imes \hat{j}) + \hat{i} \cdot (\hat{k} 	imes \hat{j})$,then $\hat{k} \cdot \hat{k} + \hat{i} \cdot (-\hat{i}) = 1 - 1 = 0$. Assuming the standard scalar triple product form $\hat{k} \cdot (\hat{i} 	imes \hat{j}) = 1$,the expression evaluates to $1 + 1 = 2$. However,if the question implies $\hat{k} \cdot (\hat{i} 	imes \hat{j}) + \hat{i} \cdot (\hat{j} 	imes \hat{k})$ is not correct,let us check $\hat{k} \cdot (\hat{i} 	imes \hat{j}) + \hat{i} \cdot (\hat{k} 	imes \hat{j}) = 1 - 1 = 0$. Given the options,the correct value is $0$.

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

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