Q68 / 180 · COMEDK 2019 · Mathematics

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 $\mathbf{a}=$

  • A
    $\hat{i}+\hat{j}$
  • B
    $\hat{\mathrm{i}}-\hat{\mathrm{k}}$
  • Correct answer
    C
    $\hat{\hat{i}}$
  • D
    $\hat{\mathrm{i}}+\hat{j}-\hat{\mathbf{i}}$

Solution

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, $$ \begin{array}{ll} & \mathbf{a} \cdot \hat{\mathbf{i}}=\mathbf{a}(\hat{\mathbf{i}}+\hat{\mathbf{j}})=\mathbf{a} \cdot(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})=1 \\ \Rightarrow & x=x+y=x+y+z=1 \\ \Rightarrow & x=1, y=0, z=0 \\ \therefore & \mathbf{a}=\hat{\mathbf{i}} \end{array} $$

Hence, the correct answer is option: 'C'.

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