Given A = hat{i} + hat{j} + hat{k} and B = -h | vedclass

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(A) Given vectors are $A = \hat{i} + \hat{j} + \hat{k}$ and $B = -\hat{i} - \hat{j} - \hat{k}$.First,calculate the vector $(A - B)$:$A - B = (\hat{i}

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(A) Given vectors are $A = \hat{i} + \hat{j} + \hat{k}$ and $B = -\hat{i} - \hat{j} - \hat{k}$.First,calculate the vector $(A - B)$:$A - B = (\hat{i} + \hat{j} + \hat{k}) - (-\hat{i} - \hat{j} - \hat{k})$$A - B = \hat{i} + \hat{j} + \hat{k} + \hat{i} + \hat{j} + \hat{k} = 2\hat{i} + 2\hat{j} + 2\hat{k}$$A - B = 2(\hat{i} + \hat{j} + \hat{k}) = 2A$Since $(A - B)$ is a positive scalar multiple of $A$,the vector $(A - B)$ is parallel to $A$.Therefore,the angle between $(A - B)$ and $A$ is $0^{\circ}$.

Given $A = \hat{i} + \hat{j} + \hat{k}$ and $B = -\hat{i} - \hat{j} - \hat{k}$,then the angle between $(A - B)$ and $A$ is: (in $^{\circ}$)

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