Verified Solution • 2025–26 Session moderate

QJEEM23Ph9352: JEE Main 2023 Physics - Class 11 JEEM23Ph9352 Solution: If and then, the unit vector in the direction of is . The value …

If P→=3i^+3j^+2k^\overrightarrow P = 3\widehat i + \sqrt 3 \widehat j + 2\widehat k and Q→=4i^+3j^+2.5k^\overrightarrow Q = 4\widehat i + \sqrt 3 \widehat j + 2.5\widehat k then, the unit vector in the direction of P→×Q→\overrightarrow P \times \overrightarrow Q is 1x(3i^+j^−23k^){1 \over x}\left( {\sqrt 3 \widehat i + \widehat j - 2\sqrt 3 \widehat k} \right) . The value of xx is _________.

Step-by-Step Verified Solution

P⃗=3i^+3j^+2k^\vec{P}=3 \hat{i}+\sqrt{3} \hat{j}+2 \hat{k}

Q⃗=4i^+3j^+2.5k^\vec{Q}=4 \hat{i}+\sqrt{3} \hat{j}+2.5 \hat{k}

P⃗×Q⃗=∣i^amp;j^amp;k^3amp;3amp;24amp;3amp;2.5∣\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|

=i^(32)−j^(−12)+k^(−3)=\hat{i}\left(\frac{\sqrt{3}}{2}\right)-\hat{j}\left(-\frac{1}{2}\right)+\hat{k}(-\sqrt{3})

=32i^+j^2−3k^=\frac{\sqrt{3}}{2} \hat{i}+\frac{\hat{j}}{2}-\sqrt{3} \hat{k}

∣P⃗×Q⃗∣=34+14+3=2|\vec{P} \times \vec{Q}|=\sqrt{\frac{3}{4}+\frac{1}{4}+3}=2

Unit vector along P⃗×Q⃗=14(3i^+j^−23k^)\vec{P} \times \vec{Q}=\frac{1}{4}(\sqrt{3} \hat{i}+\hat{j}-2 \sqrt{3} \hat{k})

x=4x=4

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