JEE 2023 — Mathematics PYQ

Solution:

First, simplify the cross product:

$(\vec{a}+2\vec{b})\times (2\vec{a}-3\vec{b})=2(\vec{a}\times \vec{a})-3(\vec{a}\times \vec{b})+4(\vec{b}\times \vec{a})-6(\vec{b}\times \vec{b})$

Since $\vec{a}\times \vec{a}=0$ and $\vec{b}\times \vec{a}=-(\vec{a}\times \vec{b})$:

$=0-3(\vec{a}\times \vec{b})-4(\vec{a}\times \vec{b})-0=-7(\vec{a}\times \vec{b})$

Now, find the magnitude squared:

$|-7(\vec{a}\times \vec{b}){|}^2=49|\vec{a}{|}^2|\vec{b}{|}^2{\sin }^2(\frac{\pi }{4})$

$=49\cdot (2{)}^2\cdot (3{)}^2\cdot (\frac{1}{\sqrt{2}}{)}^2$

$=49\cdot 4\cdot 9\cdot \frac{1}{2}=49\cdot 18=882$

Correct Option: (3) 882