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If $\mathbf{a} \cdot \hat{\mathbf{i}}=\mathbf{a} \cdot(\hat{\mathbf{i}}+\hat{\mathbf{j}})=\mathbf{a} \cdot(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})$, then $\mathbf{a}$ is equal to
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Verified Answer
The correct answer is:
$\hat{\mathbf{i}}$
We have
$$
\begin{aligned}
& \mathbf{a} \cdot \hat{\mathbf{i}}=\mathbf{a} \cdot(\hat{\mathbf{i}}+\hat{\mathbf{j}})=\mathbf{a} \cdot \hat{\mathbf{i}}+\mathbf{a} \cdot \hat{\mathbf{j}} \text { a } \hat{\mathbf{j}}=0 \\
& \text { Also, } \quad \mathbf{a} \cdot \hat{\mathbf{i}}=\mathbf{a} \cdot(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}) \\
& a \cdot \hat{\mathbf{i}}+\mathbf{a} \cdot \hat{\mathbf{j}}=\mathbf{a} \cdot \hat{\mathbf{i}}+\mathbf{a} \hat{\mathbf{j}}+\mathbf{a} \cdot \hat{\mathbf{k}} \\
& \Rightarrow \quad \hat{\mathbf{a}} \cdot \hat{\mathbf{k}}=0 \\
& \text { Also, } \quad \hat{\mathbf{i}} \cdot \hat{\mathbf{j}}=\hat{\mathbf{j}} \cdot \hat{\mathbf{k}}=\hat{\mathbf{k}} \cdot \hat{\mathbf{i}}=0 \\
& \text { Then } \mathbf{a}=\hat{\mathbf{i}} \\
&
\end{aligned}
$$
$$
\begin{aligned}
& \mathbf{a} \cdot \hat{\mathbf{i}}=\mathbf{a} \cdot(\hat{\mathbf{i}}+\hat{\mathbf{j}})=\mathbf{a} \cdot \hat{\mathbf{i}}+\mathbf{a} \cdot \hat{\mathbf{j}} \text { a } \hat{\mathbf{j}}=0 \\
& \text { Also, } \quad \mathbf{a} \cdot \hat{\mathbf{i}}=\mathbf{a} \cdot(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}) \\
& a \cdot \hat{\mathbf{i}}+\mathbf{a} \cdot \hat{\mathbf{j}}=\mathbf{a} \cdot \hat{\mathbf{i}}+\mathbf{a} \hat{\mathbf{j}}+\mathbf{a} \cdot \hat{\mathbf{k}} \\
& \Rightarrow \quad \hat{\mathbf{a}} \cdot \hat{\mathbf{k}}=0 \\
& \text { Also, } \quad \hat{\mathbf{i}} \cdot \hat{\mathbf{j}}=\hat{\mathbf{j}} \cdot \hat{\mathbf{k}}=\hat{\mathbf{k}} \cdot \hat{\mathbf{i}}=0 \\
& \text { Then } \mathbf{a}=\hat{\mathbf{i}} \\
&
\end{aligned}
$$
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