JEE Main 2025 Jan 28 Shift 1, Mathematics Q23: Vector Algebra

Step 1: $|\vec{c}|$. Expand $|\vec{c}-2\vec{a}{|}^2=|\vec{c}{|}^2-4(\vec{a}\cdot \vec{c})+4|\vec{a}{|}^2=8$. With $|\vec{a}{|}^2=3$ and $\vec{a}\cdot \vec{c}=|\vec{c}|$:

$|\vec{c}{|}^2-4|\vec{c}|+12=8\Rightarrow |\vec{c}{|}^2-4|\vec{c}|+4=0\Rightarrow |\vec{c}|=2$.

Step 2: $\vec{d}$ and $|\vec{d}|$. $\vec{d}=\vec{a}\times \vec{b}=\left|\begin{array}{c}\hat{i}\hat{j}\hat{k}\\ 111\\ 221\end{array}\right|=(1-2)\hat{i}-(1-2)\hat{j}+(2-2)\hat{k}=-\hat{i}+\hat{j}$. So $|\vec{d}|=\sqrt{2}$.

Step 3: $|\vec{d}\times \vec{c}{|}^2$. Using $|\vec{d}\times \vec{c}|=|\vec{d}||\vec{c}|\sin (\pi /4)$:

$|\vec{d}\times \vec{c}{|}^2=(\sqrt{2}{)}^2(2{)}^2\cdot \frac{1}{2}=4$.

Step 4: $\vec{b}\cdot \vec{c}$. Use the identity $|(\vec{a}\times \vec{b})\times \vec{c}{|}^2=|\vec{a}\times \vec{b}{|}^2|\vec{c}{|}^2-((\vec{a}\times \vec{b})\cdot \vec{c}{)}^2$, but it's cleaner to use $(\vec{a}\times \vec{b})\times \vec{c}=(\vec{a}\cdot \vec{c})\vec{b}-(\vec{b}\cdot \vec{c})\vec{a}$. So $\vec{d}\times \vec{c}=|\vec{c}|\vec{b}-(\vec{b}\cdot \vec{c})\vec{a}=2\vec{b}-(\vec{b}\cdot \vec{c})\vec{a}$.

Let $u=\vec{b}\cdot \vec{c}$. Then $|\vec{d}\times \vec{c}{|}^2=4|\vec{b}{|}^2-4u(\vec{a}\cdot \vec{b})+u^2|\vec{a}{|}^2$. With $|\vec{b}{|}^2=9$, $\vec{a}\cdot \vec{b}=5$, $|\vec{a}{|}^2=3$:

$4=36-20u+3u^2\Rightarrow 3u^2-20u+32=0$.

Roots: $u=\frac{20\pm \sqrt{400-384}}{6}=\frac{20\pm 4}{6}=4$ or $\frac{8}{3}$.

Step 5: combine. For each candidate $u$, compute $|10-3u|$:

If $u=4$: $|10-12|=2$, total $=2+4=6$.
If $u=8/3$: $|10-8|=2$, total $=2+4=6$.

$\therefore$ Answer $=6$.