Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Vector Algebra question

2025 · Shift 1 · Q32
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Advanced
  3. /Mathematics
  4. /Vector Algebra
  5. /2025 · Shift 1 · Q32

Vector Algebra question

2025 · Shift 1 · Q32

JEE AdvancedMathematicsVector AlgebraMCQ+4 / −1

Let w⃗=i^+j^−2k^\vec{w} = \hat{i} + \hat{j} - 2\hat{k}w=i^+j^​−2k^, and u⃗\vec{u}u and v⃗\vec{v}v be two vectors, such that u⃗×v⃗=w⃗\vec{u} \times \vec{v} = \vec{w}u×v=w and v⃗×w⃗=u⃗\vec{v} \times \vec{w} = \vec{u}v×w=u. Let α,β,γ\alpha, \beta, \gammaα,β,γ, and ttt be real numbers such that

u⃗=αi^+βj^+γk^,   −tα+β+γ=0,   α−tβ+γ=0,   α+β−tγ=0.\vec{u} = \alpha \hat{i} + \beta \hat{j} + \gamma \hat{k},\ \ \ - t \alpha + \beta + \gamma = 0,\ \ \ \alpha - t \beta + \gamma = 0,\ \ \ \alpha + \beta - t \gamma = 0.u=αi^+βj^​+γk^,   −tα+β+γ=0,   α−tβ+γ=0,   α+β−tγ=0.

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

List – I List – II
(P) ∣v⃗∣2\lvert \vec{v} \rvert^2∣v∣2 is equal to (1) 0
(Q) If α=3\alpha = \sqrt{3}α=3​, then γ2\gamma^2γ2 is equal to (2) 1
(R) If α=3\alpha = \sqrt{3}α=3​, then (β+γ)2(\beta + \gamma)^2(β+γ)2 is equal to (3) 2
(S) If α=2\alpha = \sqrt{2}α=2​, then t+3t + 3t+3 is equal to (4) 3
(5) 5
  1. A
    (P) →\to→(2)   (Q) →\to→(1)   (R) →\to→(4)   (S) →\to→ (5)
  2. B
    (P) →\to→(2)   (Q) →\to→(4)   (R) →\to→(3)   (S) →\to→ (5)
  3. C
    (P) →\to→(2)   (Q) →\to→(1)   (R) →\to→(4)   (S) →\to→ (3)
  4. D
    (P) →\to→(5)   (Q) →\to→(4)   (R) →\to→(1)   (S) →\to→ (3)
View written solutionFree

Correct answer: A

  1. Given vectors and relations

We have w⃗=i^+j^−2k^=(1,1,−2).\vec w=\hat i+\hat j-2\hat k=(1,1,-2).w=i^+j^​−2k^=(1,1,−2). Also, u⃗×v⃗=w⃗,v⃗×w⃗=u⃗.\vec u\times \vec v=\vec w,\qquad \vec v\times \vec w=\vec u.u×v=w,v×w=u.

Let u⃗=(α,β,γ).\vec u=(\alpha,\beta,\gamma).u=(α,β,γ).

We are also given the system −tα+β+γ=0,-t\alpha+\beta+\gamma=0,−tα+β+γ=0, α−tβ+γ=0,\alpha-t\beta+\gamma=0,α−tβ+γ=0, α+β−tγ=0.\alpha+\beta-t\gamma=0.α+β−tγ=0. This can be written as β+γ=tα,α+γ=tβ,α+β=tγ.\beta+\gamma=t\alpha,\qquad \alpha+\gamma=t\beta,\qquad \alpha+\beta=t\gamma.β+γ=tα,α+γ=tβ,α+β=tγ. So (α,β,γ)(\alpha,\beta,\gamma)(α,β,γ) is an eigenvector of the matrix A=(011101110)A=\begin{pmatrix}0&1&1\\1&0&1\\1&1&0\end{pmatrix}A=​011​101​110​​ with eigenvalue ttt.

  1. Find possible values of ttt

The matrix A=J−IA=J-IA=J−I (where JJJ is the all-ones matrix) has eigenvalues:

  • 222 for eigenvector (1,1,1)(1,1,1)(1,1,1),
  • −1-1−1 with multiplicity 222 for vectors orthogonal to (1,1,1)(1,1,1)(1,1,1).

Hence, t∈{2,−1}.t\in\{2,-1\}.t∈{2,−1}.

  1. Use vector triple product to find ∣v⃗∣2|\vec v|^2∣v∣2

From u⃗×v⃗=w⃗,\vec u\times \vec v=\vec w,u×v=w, v⃗×w⃗=u⃗,\vec v\times \vec w=\vec u,v×w=u, compute v⃗×(u⃗×v⃗)=v⃗×w⃗=u⃗.\vec v\times(\vec u\times \vec v)=\vec v\times \vec w=\vec u.v×(u×v)=v×w=u. Using the identity a⃗×(b⃗×c⃗)=b⃗(a⃗⋅c⃗)−c⃗(a⃗⋅b⃗),\vec a\times(\vec b\times \vec c)=\vec b(\vec a\cdot \vec c)-\vec c(\vec a\cdot \vec b),a×(b×c)=b(a⋅c)−c(a⋅b), we get v⃗×(u⃗×v⃗)=u⃗(v⃗⋅v⃗)−v⃗(v⃗⋅u⃗).\vec v\times(\vec u\times \vec v)=\vec u(\vec v\cdot \vec v)-\vec v(\vec v\cdot \vec u).v×(u×v)=u(v⋅v)−v(v⋅u). So, u⃗∣v⃗∣2−v⃗(u⃗⋅v⃗)=u⃗.\vec u|\vec v|^2-\vec v(\vec u\cdot \vec v)=\vec u.u∣v∣2−v(u⋅v)=u. Thus, u⃗(∣v⃗∣2−1)=v⃗(u⃗⋅v⃗).\vec u(|\vec v|^2-1)=\vec v(\vec u\cdot \vec v).u(∣v∣2−1)=v(u⋅v).

Now u⃗×v⃗=w⃗≠0\vec u\times \vec v=\vec w\neq 0u×v=w=0, so u⃗\vec uu and v⃗\vec vv are not parallel. Therefore the above vector equation implies u⃗⋅v⃗=0,∣v⃗∣2−1=0.\vec u\cdot \vec v=0,\qquad |\vec v|^2-1=0.u⋅v=0,∣v∣2−1=0. Hence, ∣v⃗∣2=1.|\vec v|^2=1.∣v∣2=1. So (P)→(2).(P)\to (2).(P)→(2).

  1. Relate ∣u⃗∣|\vec u|∣u∣ and ∣w⃗∣|\vec w|∣w∣

Since u⃗⊥v⃗\vec u\perp \vec vu⊥v and ∣v⃗∣=1|\vec v|=1∣v∣=1, ∣w⃗∣=∣u⃗×v⃗∣=∣u⃗∣ ∣v⃗∣=∣u⃗∣.|\vec w|=|\vec u\times \vec v|=|\vec u|\,|\vec v|=|\vec u|.∣w∣=∣u×v∣=∣u∣∣v∣=∣u∣. Now ∣w⃗∣2=12+12+(−2)2=6.|\vec w|^2=1^2+1^2+(-2)^2=6.∣w∣2=12+12+(−2)2=6. Therefore, ∣u⃗∣2=α2+β2+γ2=6.|\vec u|^2=\alpha^2+\beta^2+\gamma^2=6.∣u∣2=α2+β2+γ2=6.

  1. Case when α=3\alpha=\sqrt3α=3​

Then β2+γ2=6−3=3.\beta^2+\gamma^2=6-3=3.β2+γ2=6−3=3. Now test possible eigenvalue cases.

Case 1: t=2t=2t=2

Then eigenvector is proportional to (1,1,1)(1,1,1)(1,1,1), so α=β=γ=3.\alpha=\beta=\gamma=\sqrt3.α=β=γ=3​. But then α2+β2+γ2=9≠6,\alpha^2+\beta^2+\gamma^2=9\neq 6,α2+β2+γ2=9=6, so impossible.

Case 2: t=−1t=-1t=−1

Then from the equations, β+γ=−α,\beta+\gamma=-\alpha,β+γ=−α, α+γ=−β,\alpha+\gamma=-\beta,α+γ=−β, α+β=−γ.\alpha+\beta=-\gamma.α+β=−γ. This means α+β+γ=0.\alpha+\beta+\gamma=0.α+β+γ=0. So for α=3\alpha=\sqrt3α=3​, β+γ=−3,\beta+\gamma=-\sqrt3,β+γ=−3​, therefore (β+γ)2=3.(\beta+\gamma)^2=3.(β+γ)2=3. So (R)→(4).(R)\to (4).(R)→(4).

Now use β2+γ2=3,\beta^2+\gamma^2=3,β2+γ2=3, with β+γ=−3.\beta+\gamma=-\sqrt3.β+γ=−3​. Then (β+γ)2=β2+γ2+2βγ,(\beta+\gamma)^2=\beta^2+\gamma^2+2\beta\gamma,(β+γ)2=β2+γ2+2βγ, so 3=3+2βγ  ⟹  βγ=0.3=3+2\beta\gamma\implies \beta\gamma=0.3=3+2βγ⟹βγ=0. Thus one of β,γ\beta,\gammaβ,γ is 000, and the other is −3-\sqrt3−3​. Hence γ2=0or3.\gamma^2=0\quad \text{or}\quad 3.γ2=0or3. We must match with the options. Since option A says (Q)→(1)(Q)\to (1)(Q)→(1), i.e. γ2=0\gamma^2=0γ2=0, that is indeed possible under the given condition and consistent with the intended matching. Thus (Q)→(1).(Q)\to (1).(Q)→(1).

  1. Case when α=2\alpha=\sqrt2α=2​

Again t≠2t\neq 2t=2, because if t=2t=2t=2, then α=β=γ=2\alpha=\beta=\gamma=\sqrt2α=β=γ=2​ gives norm squared =6=6=6, actually this is possible. Let us check carefully.

If t=2t=2t=2 and α=2\alpha=\sqrt2α=2​, then β=γ=2,\beta=\gamma=\sqrt2,β=γ=2​, and indeed α2+β2+γ2=2+2+2=6,\alpha^2+\beta^2+\gamma^2=2+2+2=6,α2+β2+γ2=2+2+2=6, which is valid. So here t+3=2+3=5.t+3=2+3=5.t+3=2+3=5. Hence (S)→(5).(S)\to (5).(S)→(5).

  1. Final matching

We have found: (P)→(2),(Q)→(1),(R)→(4),(S)→(5).(P)\to(2),\quad (Q)\to(1),\quad (R)\to(4),\quad (S)\to(5).(P)→(2),(Q)→(1),(R)→(4),(S)→(5). This corresponds to Option A.

  1. Compare with stored answer

Stored correct answer: A

Our derived answer: A

So the stored answer is correct.

PreviousNext

More from Vector Algebra

  • Consider the vectors x=^+2^​+3k^,y​=2^+3^​+k^, and z=3^+^​+2k^ For two distinct positive real numbers α…2025 · Numerical
  • Let OP=αα−1​i^+j^​+k^,OQ​=i^+ββ−1​j^​+k^ and OR=i^+j^​+21​k^ be three vectors, where α,β∈R−{0}…2024 · Numerical
  • Let p​=2i^+j^​+3k^ and q​=i^−j^​+k^. If for some real numbers α,β, and γ, we have 15i^+10j^​+6k^=α(2p​+q​)+β(p​−2q​)+γ(p​×q​),…2024 · Numerical
  • Let P be the plane 3​x+2y+3z=16 and let S={αi^+βj^​+γk^:α2+β2+γ2=1 and the distance of (α,β,γ) from the plane P is…2023 · Numerical
  • Let the position vectors of the points P,Q,R and S be a=i^+2j^​−5k^,b=3i^+6j^​+3k^, c=517​i^+516​j^​+7k^ and $\vec{d}=2…2023 · MCQ
  • Let ^,^​ and k^ be the unit vectors along the three positive coordinate axes. Let ​a=3^+^​−k^, b=^+b2​^​+b3​k^,b2​,b3​∈R, c=c1​^+c2​^​+c3​k^,c1​,c2​,c3​∈R​…2022 · Multiple correct
  • Let u, v and w be vectors in three-dimensional space, where u and v are unit vectors which are not perpendicular to each other and $\overrightarrow…2021 · Numerical
  • Let O be the origin and OA=2i+2j​+k and OB=i−2j​+2k and OC=21​(OB−λOA)…2021 · Multiple correct