JEE Advanced 2025 Paper 1 · Q16 · Vector Algebra

Question

Let $\vec{w} = \hat{i} + \hat{j} - 2\hat{k}$, and let $\vec{u}, \vec{v}$ be vectors with $\vec{u} 	imes \vec{v} = \vec{w}$ and $\vec{v} 	imes \vec{w} = \vec{u}$. Let $\alpha, \beta, \gamma, t$ be reals with $\vec{u} = \alpha \hat{i} + \beta \hat{j} + \gamma \hat{k}$ and $-t\alpha + \beta + \gamma = 0$, $\alpha - t\beta + \gamma = 0$, $\alpha + \beta - t\gamma = 0$. List-I: (P) $|\vec{v}|^2$; (Q) if $\alpha = \sqrt{3}$, $\gamma^2$; (R) if $\alpha = \sqrt{3}$, $(\beta + \gamma)^2$; (S) if $\alpha = \sqrt{2}$, $t + 3$. List-II: (1) $0$; (2) $1$; (3) $2$; (4) $3$; (5) $5$. Choose: (A) P->2,Q->1,R->4,S->5; (B) P->2,Q->4,R->3,S->5; (C) P->2,Q->1,R->4,S->3; (D) P->5,Q->4,R->1,S->3.

SolveKar AI · Accepted Answer

From $\vec{v} 	imes \vec{w} = \vec{u}$, take dot with $\vec{v}$: $0 = \vec{u} \cdot \vec{v}$, so $\vec{u} \perp \vec{v}$. Also $\vec{u} \perp \vec{w}$ (since $\vec{u} = \vec{v} 	imes \vec{w}$). And $\vec{v} \perp \vec{w}$ trivially. Use BAC-CAB: $\vec{u} 	imes \vec{v} = \vec{w}$ $\Rightarrow$ $|\vec{u} 	imes \vec{v}| = |\vec{w}| = \sqrt{6}$. Apply $\vec{v} 	imes (\vec{u} 	imes \vec{v}) = \vec{u}(\vec{v} \cdot \vec{v}) - \vec{v}(\vec{v} \cdot \vec{u}) = |\vec{v}|^2 \vec{u}$ (since $\vec{u} \cdot \vec{v}=0$). But also $\vec{v} 	imes (\vec{u} 	imes \vec{v}) = \vec{v} 	imes \vec{w} = \vec{u}$, so $|\vec{v}|^2 \vec{u} = \vec{u}$, giving $|\vec{v}|^2 = 1$ (since $\vec{u} 
e \vec{0}$). (P) $	o$ $1 = (2)$. Now the linear system is the eigenvalue equation for matrix $M$ with diagonal $0$ and off-diagonal $1$; eigenvalues $2, -1, -1$. Sum the three equations: $(2 - t)(\alpha + \beta + \gamma) = 0$. Either $t = 2$ (then $\vec{u}$ along $(1,1,1)$) or $\alpha + \beta + \gamma = 0$ (giving $t = -1$). Recall $\vec{u} \perp \vec{w} = (1,1,-2)$. $\vec{u} \cdot \vec{w} = \alpha + \beta - 2\gamma = 0$. Case $t = -1$ (so $\alpha + \beta + \gamma = 0$): combined with $\alpha + \b…[truncated]

JEE Advanced 2025 Paper 1 · Q16 · Vector Algebra

Solution