Question

Let $\hat{i},\hat{j},\hat{k}$ be the orthonormal system of vectors, and $\overrightarrow{a}$ be any vector. If $\overrightarrow{a}\times \hat{i}+2\overrightarrow{a}-5\hat{j}=\overline{0}$ , then $\overrightarrow{a}$ is

• A. $2\hat{j}+\hat{k}$
• B. $2\hat{i}-\hat{k}$
• C. $2\hat{i}-\hat{j}$
• D. $2\hat{i}+\hat{k}$

Answer 1

The correct option is A $2\hat{j}+\hat{k}$

Let $\overrightarrow{a}=\lambda \hat{i}+\mu \hat{j}+\gamma \hat{k}$
$\overrightarrow{a}\times \hat{i}=-\mu \hat{k}+\gamma \hat{j}$
$\therefore$ We have $\overrightarrow{a}\times \hat{i}+2(\lambda \hat{i}+\mu \hat{j}+\gamma \hat{k})-5\hat{j}=0$
$\Rightarrow -\mu \hat{k}+\gamma \hat{j}+2\lambda \hat{i}+2\mu \hat{j}+2\gamma \hat{k}-5\hat{j}=0$
$2\mu +\gamma =5\cdots \left(1\right)$
$\mu =2\gamma \cdots \left(2\right)$
$\lambda =0\cdots \left(3\right)$
From $\left(1\right)$ and $\left(2\right)$
$\gamma =1,\mu =2$
Thus $\overrightarrow{a}=2\hat{j}+\hat{k}$