If a cdot hat{i} = a cdot (hat{i} + hat{j}) = | vedclass

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(A) Let $a = x\hat{i} + y\hat{j} + z\hat{k}$.Given $a \cdot \hat{i} = a \cdot (\hat{i} + \hat{j}) = a \cdot (\hat{i} + \hat{j} + \hat{k})$.From $a \cd

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$\hat{i}$

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(A) Let $a = x\hat{i} + y\hat{j} + z\hat{k}$.Given $a \cdot \hat{i} = a \cdot (\hat{i} + \hat{j}) = a \cdot (\hat{i} + \hat{j} + \hat{k})$.From $a \cdot \hat{i} = a \cdot (\hat{i} + \hat{j})$, we have $a \cdot \hat{i} = a \cdot \hat{i} + a \cdot \hat{j}$, which implies $a \cdot \hat{j} = 0$. Thus, $y = 0$.From $a \cdot (\hat{i} + \hat{j}) = a \cdot (\hat{i} + \hat{j} + \hat{k})$, we have $a \cdot (\hat{i} + \hat{j}) = a \cdot (\hat{i} + \hat{j}) + a \cdot \hat{k}$, which implies $a \cdot \hat{k} = 0$. Thus, $z = 0$.Since $a \cdot \hat{i} = x$, and the problem does not specify the magnitude, if we assume $a$ is a unit vector or simply look for the component form, $a = x\hat{i}$. Given the options, $a = \hat{i}$ satisfies the conditions.

If $a \cdot \hat{i} = a \cdot (\hat{i} + \hat{j}) = a \cdot (\hat{i} + \hat{j} + \hat{k})$, then $a$ is equal to

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