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 \[ \vec{u}=\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k},-t \alpha+\beta+\gamma=0, \alpha-t \beta+\gamma=0, \text { and } \alpha+\beta-t \gamma=0 . \] 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 |

(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)

Vector Algebra — JEE Advanced · 2025 · 18 May 2025 · Paper 1 Question with Solution

Let

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

, and

u

and

v

be two vectors such that

u \times v = w

and

v \times w = u

. Let

α, β, γ

and

t

be real numbers such that

u = α \hat{i} + β \hat{j} + γ \hat{k}, - t α + β + γ = 0, α - tβ + γ = 0, and α + β - t γ = 0.

Match each entry in List-I to the correct entry in List-II and choose the correct option.

Answer: (A)

From

u \times v = w

and

v \times w = u

, we have

u ⊥ v

u ⊥ w

, and

v ⊥ w

.

From

v \times w = u

∣ v ∣∣ w ∣ = ∣ u ∣

From

u \times v = w

∣ u ∣∣ v ∣ = ∣ w ∣

Multiplying:

∣ v ∣^{2} ∣ w ∣ = ∣ w ∣

, so

∣ v ∣^{2} = 1

. (P)→(2)

Also,

∣ u ∣ = ∣ w ∣ = 6

.

The system of equations gives determinant:

- t 11 1 - t 1 11 - t = 0

This gives

t = - 1

t = 2

.

Case 1:

t = 2

gives

α = β = γ

. With

α^{2} + β^{2} + γ^{2} = 6

, we get

α = \pm 2

.

For

α = 2

t + 3 = 2 + 3 = 5

. (S)→(5)

Case 2:

t = - 1

gives

α + β + γ = 0

and

α + β - 2 γ = 0

, so

γ = 0

and

α = - β

.

For

α = 3

β = - 3

γ = 0

.

(Q)

γ^{2} = 0

→ (1)

(R)

(β + γ)^{2} = (- 3)^{2} = 3

→ (4)

Answer: (P)→(2), (Q)→(1), (R)→(4), (S)→(5)

JEE Advanced · 2025 · 18 May 2025 · Paper 1 · Single correct option