Let w = i + j -2 k , and u and v be two vectors such that u × v = w and v × w = u . Let α, β, γ and t be real numbers such that u =α i +β j +γ k ,-t α+β+γ=0, α-t β+γ=0 , and α+β-t γ=0 . Match each entry in List-I to the

Correct answer: A. (P) arrow(2),(Q) arrow(1),(R) arrow(4),(S) arrow(5) array lll u × v = w & u v & u w v × w = u & v w & array aligned & w = i + j -2 k & u =α i +β j +γ k aligned | array ccc -t & 1 & 1 1 & -t & 1 1 & 1 & -t array |=0 array ll u . w =0=α+β-2 γ & t=-1 or 2 ( v . w ) × v = u × v & t=-1 w | v |^2-0= w & α+β+γ=0 | v |=1(P) & t=2 & α=β=γ array aligned & Also, u × v = w & | u || v |=| w | & | u |=| w |= 6 aligned Case-1: aligned & t=2 & α=β=γ & α^2+β^2+γ^2=6 & α= 2 ,- 2 & (S) t+3=5 aligned Case-2: aligned & t=-1 & α+β+γ=0 & α+β-2 γ=0 & γ=0, α=-β & (Q) α= 3 , γ^2=0, β=- 3 & (R) α= 3 (γ+β)^2=β^2=3 & (P) arrow(2),(Q) arrow(1),(R) arrow(4),(S) arrow(5) aligned

JEE Advanced 2025Paper 1

Exam: JEE Advanced
Year: 2025
Paper: Paper 1
Subject: Mathematics
Chapter: Vector Algebra

JEE Advanced 2025 · Paper 1 · Mathematics · Vector Algebra

Let w = i + j -2 k , and u and v be two vectors such that u × v = w...

Let

\vec{w}=\hat{i}+\hat{j}-2 \hat{k}

, and

\overrightarrow{\mathrm{u}}

and

\overrightarrow{\mathrm{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.

	LIST - I		LIST - II
(P)	$\|\vec{v}\|^2$ is equal to	(1)	0
(Q)	If $\alpha=\sqrt{3}$ ,then $\gamma^2$ is equal to	(2)	1
(R)	If $\alpha=\sqrt{3}$ ,then $(\beta+\gamma)^2$ is equal to	(3)	2
(S)	If $\alpha=\sqrt{2}$ ,then $t+3$ is equal to	(4)	3
		(5)	5

Options

A $(\mathrm{P}) \rightarrow(2),(\mathrm{Q}) \rightarrow(1),(\mathrm{R}) \rightarrow(4),(\mathrm{S}) \rightarrow(5)$
B $(\mathrm{P}) \rightarrow(2),(\mathrm{Q}) \rightarrow(4),(\mathrm{R}) \rightarrow(3),(\mathrm{S}) \rightarrow(5)$
C $(\mathrm{P}) \rightarrow(2),(\mathrm{Q}) \rightarrow(1),(\mathrm{R}) \rightarrow(4),(\mathrm{S}) \rightarrow(3)$
D $(\mathrm{P}) \rightarrow(5),(\mathrm{Q}) \rightarrow(4),(\mathrm{R}) \rightarrow(1),(\mathrm{S}) \rightarrow(3)$

Correct answer

A. $(\mathrm{P}) \rightarrow(2),(\mathrm{Q}) \rightarrow(1),(\mathrm{R}) \rightarrow(4),(\mathrm{S}) \rightarrow(5)$

Step-by-step solution

\begin{array}{lll}\vec{u} \times \vec{v}=\vec{w} & \vec{u} \perp \vec{v} & \vec{u} \perp \vec{w} \\ \vec{v} \times \vec{w}=\vec{u} & \vec{v} \perp \vec{w} & \end{array}

\begin{aligned} & \overrightarrow{\mathrm{w}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{u}}=\alpha \hat{\mathrm{i}}+\beta \hat{\mathrm{j}}+\gamma \hat{\mathrm{k}}\end{aligned}

\quad\left|

\begin{array}{ccc}-\mathrm{t} & 1 & 1 \\ 1 & -\mathrm{t} & 1 \\ 1 & 1 & -\mathrm{t}\end{array}

\right|=0

\begin{array}{ll}\vec{u} . \vec{w}=0=\alpha+\beta-2 \gamma & t=-1 \text{ or } 2 \\ (\vec{v} . \vec{w}) \times \vec{v}=\vec{u} \times \vec{v} & t=-1 \\ \vec{w}|\vec{v}|^2-0=\vec{w} & \alpha+\beta+\gamma=0 \\ |\vec{v}|=1(P) & t=2 \\ & \alpha=\beta=\gamma\end{array}

\begin{aligned} & \text{ Also, } \vec{u} \times \vec{v}=\vec{w} \\ & |\vec{u}||\vec{v}|=|\vec{w}| \\ & \Rightarrow|\vec{u}|=|\vec{w}|=\sqrt{6}\end{aligned}

Case-1:

\begin{aligned} & \mathrm{t}=2 \\ & \alpha=\beta=\gamma \\ & \alpha^2+\beta^2+\gamma^2=6 \\ & \alpha=\sqrt{2},-\sqrt{2} \\ & \text{ (S) } \mathrm{t}+3=5 \end{aligned}

Case-2:

\begin{aligned} & \mathrm{t}=-1 \\ & \alpha+\beta+\gamma=0 \\ & \alpha+\beta-2 \gamma=0 \\ & \gamma=0, \alpha=-\beta \\ & (\mathrm{Q}) \alpha=\sqrt{3}, \gamma^2=0, \beta=-\sqrt{3} \\ & (\mathrm{R}) \alpha=\sqrt{3}(\gamma+\beta)^2=\beta^2=3 \\ & (\mathrm{P}) \rightarrow(2),(\mathrm{Q}) \rightarrow(1),(\mathrm{R}) \rightarrow(4),(\mathrm{S}) \rightarrow(5) \end{aligned}$$

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