Question

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

• A. $\hat{i}+\hat{j}$
• B. $\hat{i}-\hat{k}$
• C. $\hat{i}$
• D. $\hat{i}+\hat{j}-\hat{k}$

Answer 1

Answer: (C)

$\hat{i}$

Let $\overrightarrow{a}=a_1\hat{i}+a_2\hat{j}+a_3\overrightarrow{k}$

So given $\overrightarrow{a}.\hat{i}=1$

$\Rightarrow (a_1\hat{i}+a_2\overrightarrow{j}+a_3\overrightarrow{k}).\overrightarrow{i}=1$

$\Rightarrow a_1=1$

$a.(\hat{i}+\hat{j})=1$

$\Rightarrow (a_1\hat{i}+a_2\hat{j}+a_3\hat{k}).(\hat{i}+\hat{j})=1$

$\Rightarrow a_1+a_2=1$

$a.(\hat{i}+\hat{j}+\hat{k})=1$

$\Rightarrow (a_1\hat{i}+a_2\hat{j}+a_3\hat{k}).(\hat{i}+\hat{j}+\hat{k})=1$

$\Rightarrow a_1+a_2+a_3=1$

So, from $3$ equations we get, $a_1=1,a_2=0,a_3=0$

$\therefore \overrightarrow{a}=1\hat{i}+0\hat{j}+0\hat{k}=\hat{i}$