JEE Advanced 2025 Paper 1Question 1
Let $\vec w=\hat i+\hat j-2\hat k$, and $\vec u$ and $\vec v$ be two vectors, such that $\vec u\times\vec v=\vec w$ and $\vec v\times\vec w=\vec u$. Let $\alpha,\beta,\gamma$, and $t$ be real numbers such that $\vec u=\alpha\hat i+\beta\hat j+\gamma\hat k$, $-t\alpha+\beta+\gamma=0$, $\alpha-t\beta+\gamma=0$, and $\alpha+\beta-t\gamma=0$.\\ Match each entry in List-I to the correct entry in List-II and choose the correct option.\\ $\begin{array}{ll}\textbf{List-I} & \textbf{List-II}\\ \text{(P) } |\vec v|^2 \text{ is equal to} & \text{(1) } 0\\ \text{(Q) If } \alpha=\sqrt3, \text{ then } \gamma^2 \text{ is equal to} & \text{(2) } 1\\ \text{(R) If } \alpha=\sqrt3, \text{ then } (\beta+\gamma)^2 \text{ is equal to} & \text{(3) } 2\\ \text{(S) If } \alpha=\sqrt2, \text{ then } t+3 \text{ is equal to} & \text{(4) } 3\\ & \text{(5) } 5\end{array}$
Vector Algebra
Solution
$\vec u\perp\vec v\perp\vec w$ with $|\vec u|=|\vec v||\vec w|$ and $|\vec w|=|\vec u||\vec v|$ gives $|\vec v|^2=1$, $|\vec u|^2=6$. $\vec u\cdot\vec w=0$ and the linear system force $\vec u\parallel(1,1,1)$ ($t=2$) or $\alpha+\beta+\gamma=0$ ($t=-1$). $\alpha=\sqrt3$: $\vec u=(\sqrt3,-\sqrt3,0)$, $\gamma^2=0$, $(\beta+\gamma)^2=3$. $\alpha=\sqrt2$: $\vec u=\sqrt2(1,1,1)$, $t=2$, $t+3=5$. So P$\to$2, Q$\to$1, R$\to$4, S$\to$5.