Q83 / 180 · KCET 2008 · Mathematics

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

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

Solution

Let $\overrightarrow{\mathbf{a}}=\mathrm{a}_{1} \hat{\mathbf{i}}+\mathrm{a}_{2} \hat{\mathbf{j}}+\mathrm{a}_{3} \hat{\mathbf{k}}$ Given, $\overrightarrow{\mathbf{a}} \cdot \hat{\mathbf{i}}=\overrightarrow{\mathbf{a}} \cdot(\hat{\mathbf{i}}+\hat{\mathbf{j}})=\overrightarrow{\mathbf{a}} \cdot(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})=1$ $\therefore \quad \mathrm{a}_{1}=\mathrm{a}_{1}+\mathrm{a}_{2}=\mathrm{a}_{1}+\mathrm{a}_{2}+\mathrm{a}_{3}=1$ $\Rightarrow \quad \mathrm{a}_{1}=1, \mathrm{a}_{2}=0, \mathrm{a}_{3}=0$ $\therefore \quad \overrightarrow{\mathbf{a}}=\hat{\mathbf{i}}$

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

Relevant Questions