Question

The vector $\overrightarrow{OP}=\hat{i}+\hat{j}+\hat{k}$ is turned through a right angle about the origin $O$ so that it passes through the positive side of $y-$axis.Its new position is

• A. $\sqrt{\frac{3}{2}}(\hat{i}-\hat{j})$
• B. $\sqrt{\frac{3}{2}}(\hat{j}-\hat{k})$
• C. $\sqrt{\frac{3}{2}}(\hat{i}-\hat{k})$
• D. $\frac{1}{\sqrt{2}}(-\hat{i}+2\hat{j}-\hat{k})$

Answer 1

The correct option is D $\frac{1}{\sqrt{2}}(-\hat{i}+2\hat{j}-\hat{k})$

Let the new vector be,$\overrightarrow{OQ}=\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}$
Given $\overrightarrow{OP},\overrightarrow{OQ},\hat{j}$ are coplanar.
$\therefore \left|\begin{array}{ccc}1& 1& 1\\ \alpha & \beta & \gamma \\ 0& 1& 0\end{array}\right|=0$
$\Rightarrow 1(0-\gamma )-1\left(0\right)+1\left(\alpha \right)=0$
$\Rightarrow \gamma =\alpha$ ..........$\left(1\right)$
$\overrightarrow{OP}\perp \overrightarrow{OQ}$
$\Rightarrow \alpha +\beta +\gamma =0,\beta =-2\gamma$ ................. $\left(2\right)$
$\left|\overrightarrow{OQ}\right|=\left|\overrightarrow{OP}\right|\Rightarrow {\alpha }^2+{\beta }^2+{\gamma }^2=3$
$\alpha =\pm \frac{1}{\sqrt{2}}$ where $\alpha =\gamma$ from $\left(1\right)$
$\Rightarrow \overrightarrow{OQ}=\frac{-1}{\sqrt{2}}(\hat{i}-2\hat{j}+\hat{k})$
where coefficient of $\hat{j}>0$
$\therefore$ its new position is at $\frac{-1}{\sqrt{2}}(\hat{i}-2\hat{j}+\hat{k})$