Verified Solution • 2025–26 Session moderate

QJEEM23Ma3f85: JEE Main 2023 Physics - Class 11 JEEM23Ma3f85 Solution: Let and the angle between the vectors and be . Then is equal to :

Let ∣a⃗∣=2,∣b⃗∣=3|\vec{a}|=2,|\vec{b}|=3 and the angle between the vectors a⃗\vec{a} and b⃗\vec{b} be π4\frac{\pi}{4} . Then ∣(a⃗+2b⃗)×(2a⃗−3b⃗)∣2|(\vec{a}+2 \vec{b}) \times(2 \vec{a}-3 \vec{b})|^{2} is equal to :

Multiple Choice Options

  • A 441
  • B 482
  • C 841
  • D 882

Step-by-Step Verified Solution

amp;∣a⃗∣=2amp;∣b⃗∣=3amp;a⃗⋅b⃗=π4\begin{aligned}& |\vec{a}|=2 \\\\& |\vec{b}|=3 \\\\& \vec{a} \cdot \vec{b}=\frac{\pi}{4}\end{aligned}

amp;∣(a⃗+2b⃗)×(2a⃗−3b⃗)∣2amp;=∣−3a⃗×b⃗+4b⃗×a⃗∣2amp;=∣−3a⃗×b⃗−4a⃗×b⃗∣2amp;=∣−7a⃗×b⃗∣2amp;=(−7∣a⃗∣×∣b⃗∣sin⁡(π4))2\begin{aligned}& |(\vec{a}+2 \vec{b}) \times(2 \vec{a}-3 \vec{b})|^2 \\\\& = |-3 \vec{a} \times \vec{b}+4 \vec{b} \times \vec{a}|^2 \\\\& = |-3 \vec{a} \times \vec{b}-4 \vec{a} \times \vec{b}|^2 \\\\& = |-7 \vec{a} \times \vec{b}|^2 \\\\& = \left(-7|\vec{a}| \times|\vec{b}| \sin \left(\frac{\pi}{4}\right)\right)^2\end{aligned}

= 49×4×9×12=88249 \times 4 \times 9 \times \frac{1}{2}=882

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