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Question: Answered & Verified by Expert
If $\mathbf{a} \hat{\mathfrak{i}}=\mathbf{a}(\hat{\hat{i}}+\hat{\hat{j}})=\mathbf{a}(\hat{\mathbf{i}}+\hat{\hat{b}}+\hat{\mathbf{k}})=1$, then $\mathbf{a}=$
MathematicsVector AlgebraCOMEDKCOMEDK 2019
Options:
  • A $\hat{i}+\hat{j}$
  • B $\hat{\mathrm{i}}-\hat{\mathrm{k}}$
  • C $\hat{\hat{i}}$
  • D $\hat{\mathrm{i}}+\hat{j}-\hat{\mathbf{i}}$
Solution:
1685 Upvotes Verified Answer
The correct answer is: $\hat{\hat{i}}$
Let $\mathbf{a}=x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}$
Now, $\mathbf{a} \cdot \hat{\mathbf{i}}=(x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}) \cdot \hat{\mathbf{i}}=x$
$$
\mathbf{a} \cdot(\hat{\mathbf{i}}+\hat{\mathbf{j}})=(x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}) \cdot(\hat{\mathbf{i}}+\hat{\mathbf{j}})=x+y
$$
and $\mathbf{a} \cdot(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})=(x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}) \cdot(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})=x+y+z$
Now, according to the question,
$$
\begin{array}{ll}
& \mathbf{a} \cdot \hat{\mathbf{i}}=\mathbf{a}(\hat{\mathbf{i}}+\hat{\mathbf{j}})=\mathbf{a} \cdot(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})=1 \\
\Rightarrow & x=x+y=x+y+z=1 \\
\Rightarrow & x=1, y=0, z=0 \\
\therefore & \mathbf{a}=\hat{\mathbf{i}}
\end{array}
$$

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