Q78 / 160 · TS EAMCET 2006 · Mathematics
$\overrightarrow{\mathbf{a}} \cdot \hat{\mathbf{i}}=\overrightarrow{\mathbf{a}} \cdot(2 \hat{\mathbf{i}}+\hat{\mathbf{j}})=\overrightarrow{\mathbf{a}}(\hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}})=1$, then $\overrightarrow{\mathbf{a}}$ is equal to :
- A$\hat{\mathbf{i}}-\hat{\mathbf{k}}$
- B$1 / 3(3 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+\hat{\mathbf{k}})$
- C$1 / 3(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})$
- Correct answerD$1 / 3(3 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+\hat{\mathbf{k}})$
Solution
Let $\overrightarrow{\mathbf{a}}=a_1 \hat{\mathbf{i}}+a_2 \hat{\mathbf{j}}+a_3 \hat{\mathbf{k}}$
$\because \quad \overrightarrow{\mathbf{a}} \cdot \hat{\mathbf{i}}=1$
$\therefore \quad\left(a_1 \hat{\mathbf{i}}+a_2 \hat{\mathbf{j}}+a_3 \hat{\mathbf{k}}\right) \cdot \hat{\mathbf{i}}=1$
$\Rightarrow \quad a_1=1$
Now, $\quad \overrightarrow{\mathbf{a}} \cdot(2 \hat{\mathbf{i}}+\hat{\mathbf{j}})=1$
$\Rightarrow \quad\left(a_1 \hat{\mathbf{i}}+a_2 \hat{\mathbf{j}}+a_3 \hat{\mathbf{k}}\right) \cdot(2 \hat{\mathbf{i}}+\hat{\mathbf{j}})=1$
$\Rightarrow \quad 2 a_1+a_2=1$
$\Rightarrow \quad a_2=1-2 \quad\left(\because a_1=1\right)$
$\Rightarrow \quad a_2=-1$
and $\quad \overrightarrow{\mathbf{a}} \cdot(\hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}})=1$
$\Rightarrow \quad\left(a_1 \hat{\mathbf{i}}+a_2 \hat{\mathbf{j}}+a_3 \hat{\mathbf{k}}\right) \cdot(\hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}})=1$
$\Rightarrow \quad a_1+a_2+3 a_3=1$
$\Rightarrow \quad 1-1+3 a_3=1$
$\Rightarrow \quad a_3=\frac{1}{3}$
$\therefore \overrightarrow{\mathbf{a}}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+\frac{1}{3} \hat{\mathbf{k}}=\frac{1}{3}(3 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+\hat{\mathbf{k}})$
Hence, the correct answer is option: 'D'.