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(D) Let $\overrightarrow{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}$.Given $\overrightarrow{a} \cdot \hat{i} = 1$,we have $(a_1 \hat{i} + a_2 \hat{j

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$\frac{1}{3}(3 \hat{i} - 3 \hat{j} + \hat{k})$

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(D) Let $\overrightarrow{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}$.Given $\overrightarrow{a} \cdot \hat{i} = 1$,we have $(a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}) \cdot \hat{i} = 1$,which implies $a_1 = 1$.Next,$\overrightarrow{a} \cdot (2 \hat{i} + \hat{j}) = 1$,so $(a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}) \cdot (2 \hat{i} + \hat{j}) = 1$.This gives $2a_1 + a_2 = 1$. Substituting $a_1 = 1$,we get $2(1) + a_2 = 1$,so $a_2 = -1$.Finally,$\overrightarrow{a} \cdot (\hat{i} + \hat{j} + 3 \hat{k}) = 1$,so $(a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}) \cdot (\hat{i} + \hat{j} + 3 \hat{k}) = 1$.This gives $a_1 + a_2 + 3a_3 = 1$. Substituting $a_1 = 1$ and $a_2 = -1$,we get $1 - 1 + 3a_3 = 1$,so $3a_3 = 1$,which means $a_3 = \frac{1}{3}$.Thus,$\overrightarrow{a} = \hat{i} - \hat{j} + \frac{1}{3} \hat{k} = \frac{1}{3}(3 \hat{i} - 3 \hat{j} + \hat{k})$.

If $\overrightarrow{a} \cdot \hat{i} = \overrightarrow{a} \cdot (2 \hat{i} + \hat{j}) = \overrightarrow{a} \cdot (\hat{i} + \hat{j} + 3 \hat{k}) = 1$,then $\overrightarrow{a}$ is equal to :

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