Search any question & find its solution
Question:
Answered & Verified by Expert
$\overrightarrow{\mathbf{a}} \cdot \hat{\mathbf{i}}=\overrightarrow{\mathbf{a}} \cdot(2 \hat{\mathbf{i}}+\hat{\mathbf{j}})=\overrightarrow{\mathbf{a}}(\hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}})=1$, then $\overrightarrow{\mathbf{a}}$ is equal to :
Options:
Solution:
2586 Upvotes
Verified Answer
The correct answer is:
$1 / 3(3 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+\hat{\mathbf{k}})$
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 $\quad \overrightarrow{\mathbf{a}} \cdot(\hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}})=1$
$\Rightarrow \quad\left(a_1 \hat{\mathbf{i}}+a_2 \hat{\mathbf{j}}+a_3 \hat{\mathbf{k}}\right) \cdot(\hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}})=1$
$\Rightarrow \quad a_1+a_2+3 a_3=1$
$\Rightarrow \quad 1-1+3 a_3=1$
$\Rightarrow \quad a_3=\frac{1}{3}$
$\therefore \overrightarrow{\mathbf{a}}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+\frac{1}{3} \hat{\mathbf{k}}=\frac{1}{3}(3 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+\hat{\mathbf{k}})$
$\because \quad \overrightarrow{\mathbf{a}} \cdot \hat{\mathbf{i}}=1$
$\therefore \quad\left(a_1 \hat{\mathbf{i}}+a_2 \hat{\mathbf{j}}+a_3 \hat{\mathbf{k}}\right) \cdot \hat{\mathbf{i}}=1$
$\Rightarrow \quad a_1=1$
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 $\quad \overrightarrow{\mathbf{a}} \cdot(\hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}})=1$
$\Rightarrow \quad\left(a_1 \hat{\mathbf{i}}+a_2 \hat{\mathbf{j}}+a_3 \hat{\mathbf{k}}\right) \cdot(\hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}})=1$
$\Rightarrow \quad a_1+a_2+3 a_3=1$
$\Rightarrow \quad 1-1+3 a_3=1$
$\Rightarrow \quad a_3=\frac{1}{3}$
$\therefore \overrightarrow{\mathbf{a}}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+\frac{1}{3} \hat{\mathbf{k}}=\frac{1}{3}(3 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+\hat{\mathbf{k}})$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.