JEE MAIN - Physics (2023 - 25th January Morning Shift - No. 17)
If $$\overrightarrow P = 3\widehat i + \sqrt 3 \widehat j + 2\widehat k$$ and $$\overrightarrow Q = 4\widehat i + \sqrt 3 \widehat j + 2.5\widehat k$$ then, the unit vector in the direction of $$\overrightarrow P \times \overrightarrow Q $$ is $${1 \over x}\left( {\sqrt 3 \widehat i + \widehat j - 2\sqrt 3 \widehat k} \right)$$. The value of $$x$$ is _________.
Answer
4
Explanation
$\vec{P}=3 \hat{i}+\sqrt{3} \hat{j}+2 \hat{k}$
$\vec{Q}=4 \hat{i}+\sqrt{3} \hat{j}+2.5 \hat{k}$
$\vec{P} \times \vec{Q}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 3 & \sqrt{3} & 2 \\ 4 & \sqrt{3} & 2.5\end{array}\right|$
$=\hat{i}\left(\frac{\sqrt{3}}{2}\right)-\hat{j}\left(-\frac{1}{2}\right)+\hat{k}(-\sqrt{3})$
$=\frac{\sqrt{3}}{2} \hat{i}+\frac{\hat{j}}{2}-\sqrt{3} \hat{k}$
$|\vec{P} \times \vec{Q}|=\sqrt{\frac{3}{4}+\frac{1}{4}+3}=2$
Unit vector along $\vec{P} \times \vec{Q}=\frac{1}{4}(\sqrt{3} \hat{i}+\hat{j}-2 \sqrt{3} \hat{k})$
$x=4$
$\vec{Q}=4 \hat{i}+\sqrt{3} \hat{j}+2.5 \hat{k}$
$\vec{P} \times \vec{Q}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 3 & \sqrt{3} & 2 \\ 4 & \sqrt{3} & 2.5\end{array}\right|$
$=\hat{i}\left(\frac{\sqrt{3}}{2}\right)-\hat{j}\left(-\frac{1}{2}\right)+\hat{k}(-\sqrt{3})$
$=\frac{\sqrt{3}}{2} \hat{i}+\frac{\hat{j}}{2}-\sqrt{3} \hat{k}$
$|\vec{P} \times \vec{Q}|=\sqrt{\frac{3}{4}+\frac{1}{4}+3}=2$
Unit vector along $\vec{P} \times \vec{Q}=\frac{1}{4}(\sqrt{3} \hat{i}+\hat{j}-2 \sqrt{3} \hat{k})$
$x=4$
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