Mathematics in Physics

JEE Main level, numerical answer, medium

43 of 121

Question 43 of 121

If P⃗=3i^+3j^+2k^\vec{P}=3\hat{i}+\sqrt{3}\hat{j}+2\hat{k} and Q⃗=4i^+3j^+2.5k^\vec{Q}=4\hat{i}+\sqrt{3}\hat{j}+2.5\hat{k} then, the unit vector in the direction of P⃗×Q⃗\vec{P}\times \vec{Q} is 1x(3i^+j^−23k^)\dfrac{1}{x}(\sqrt{3}\hat{i}+\hat{j}-2\sqrt{3}\hat{k}). The value of xx is

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Answer: 4

Unit vector in the direction of P⃗×Q⃗isn^=P⃗×Q⃗∣P⃗×Q⃗∣\vec{P}\times \vec{Q}\text{is}\hat{n}=\dfrac{\vec{P}\times \vec{Q}}{|\vec{P}\times \vec{Q}|}.
Here, P⃗×Q⃗=∣i^j^k^332432.5∣=3i^2+j^2−3k^\vec{P}\times \vec{Q}=\left|\begin{matrix}\hat{i} & \hat{j} & \hat{k} \\ 3 & \sqrt{3} & 2 \\ 4 & \sqrt{3} & 2.5\end{matrix}\right|=\sqrt{3}\dfrac{\hat{i}}{2}+\dfrac{\hat{j}}{2}-\sqrt{3}\hat{k}
And ∣P⃗×Q⃗∣=(32)2+(12)2+(3)2=4=2\left|\vec{P}\times \vec{Q}\right|=\sqrt{{\left(\dfrac{\sqrt{3}}{2}\right)}^{2}+{\left(\dfrac{1}{2}\right)}^{2}+{\left(\sqrt{3}\right)}^{2}}=\sqrt{4}=2
⇒P⃗×Q⃗∣P⃗×Q⃗∣=12(3i^2+j^2−3k^)\Rightarrow \dfrac{\vec{P}\times \vec{Q}}{|\vec{P}\times \vec{Q}|}=\dfrac{1}{2}\left(\sqrt{3}\dfrac{\hat{i}}{2}+\dfrac{\hat{j}}{2}-\sqrt{3}\hat{k}\right)
=14(3i^+j^−23k^)⇒x=4=\dfrac{1}{4}(\sqrt{3}\hat{i}+\hat{j}-2\sqrt{3}\hat{k}) \Rightarrow x=4

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