NDA Vector Algebra previous year questions with solutions

Answer: (b) 1 : 3

Method. rewrite the relation as a section formula for b⃗

3a⃗ - 4b⃗ + c⃗ = 0⃗ gives b⃗ = 3a⃗ + 1·c⃗4.
Then AB⃗ = b⃗ - a⃗ = 14(c⃗ - a⃗) and BC⃗ = c⃗ - b⃗ = 34(c⃗ - a⃗).
So A, B, C are collinear with B between them, and AB: BC = 1 : 3.

Answer: (a) −⁠4

Method. in a regular hexagon each long diagonal is parallel to two sides and twice as long

Take side 1 and centre O at the origin: A(1, 0), B(12, ⁠√32),⁠ C(-⁠12⁠, ⁠√32⁠), D(-1, 0), E(-⁠12⁠, -⁠√32⁠), F(12, -⁠√32).⁠
AD⃗ = (-2, 0), BC⃗ = (-1, 0), so AD⃗ = 2BC⃗: m = 2.
CF⃗ = (1, -⁠√3),⁠ AB⃗ = (-12, ⁠√32),⁠ so CF⃗ = -2AB⃗: n = -2.
mn = -4.

Answer: (c) β = -40

Method. points with given position vectors are collinear ⇔ AB⃗ ∥ AC⃗

Read the vectors as position vectors of A(60, 3), B(40, −⁠8), C(β, −⁠52) (the first two vectors are not parallel, so it is the three points that are collinear).
AB⃗ = −⁠20î − 11ĵ, AC⃗ = (β − 60)î − 55ĵ.
Parallel ⇒ β − 60−⁠20 = −⁠55−⁠11 = 5 ⇒ β − 60 = −⁠100 ⇒ β = −⁠40.

Answer: (d) a⃗ - b⃗2

Method. diagonals of a parallelogram: PR⃗ = PQ⃗ + QR⃗, QS⃗ = QR⃗ − PQ⃗

In parallelogram PQRS, PS⃗ = QR⃗.
PR⃗ = PQ⃗ + QR⃗ = a⃗; QS⃗ = QP⃗ + PS⃗ = −⁠PQ⃗ + QR⃗ = b⃗.
Subtract: 2PQ⃗ = a⃗ − b⃗ ⇒ PQ⃗ = a⃗ − b⃗2.

Answer: (c) 83, 23

Method. collinear vectors have proportional components

pî + qĵ − 2k̂ = λ(4î + ĵ − 3k̂).
From the k̂ components: −⁠2 = −⁠3λ ⇒ λ = 23.
p = 4λ = 83, q = λ = 23.

Answer: (c) b⃗ + c⃗ − 2a⃗3

Method. centroid g⃗ = a⃗ + b⃗ + c⃗3; AG⃗ = g⃗ − a⃗

Position vector of G: a⃗ + b⃗ + c⃗3.
AG⃗ = a⃗ + b⃗ + c⃗3 − a⃗ = b⃗ + c⃗ − 2a⃗3.

Answer: (a) One

Method. triangle law: AB⃗ + BC⃗ = AC⃗ = -CA⃗

1. AB⃗ + BC⃗ + CA⃗ = AC⃗ + CA⃗ = 0⃗: correct.
2. AB⃗ + BC⃗ - CA⃗ = AC⃗ + AC⃗ = 2AC⃗ ≠ 0⃗: wrong.
3. AB⃗ + CA⃗ - BC⃗ = CB⃗ + CB⃗ = 2CB⃗ ≠ 0⃗: wrong.
4. BA⃗ - BC⃗ + CA⃗ = CA⃗ + CA⃗ = 2CA⃗ ≠ 0⃗ (since BA⃗ - BC⃗ = CA⃗): wrong.
Only one statement is correct.

Answer: (b) 4OP⃗

Method. diagonals of a parallelogram bisect each other

P is the midpoint of AC and of BD.
OA⃗ + OC⃗ = 2OP⃗ and OB⃗ + OD⃗ = 2OP⃗.
Adding: OA⃗ + OB⃗ + OC⃗ + OD⃗ = 4OP⃗.

Answer: (b) BA⃗ + CD⃗ = BD⃗ + CA⃗

Method. write every vector as a difference of position vectors

Let A, B, C, D have position vectors a⃗, b⃗, c⃗, d⃗.
BA⃗ + CD⃗ = (a⃗ - b⃗) + (d⃗ - c⃗).
BD⃗ + CA⃗ = (d⃗ - b⃗) + (a⃗ - c⃗), the same sum, so (b) holds.
Check: (a) gives -(BA⃗ + CD⃗); (c) gives c⃗ + d⃗ - a⃗ - b⃗; (d) gives c⃗ + d⃗ - a⃗ - b⃗; none equals it in general.

Answer: (c) π/⁠4

Method. cos of angle with z-axis = z-component|r⃗|

|r⃗| = ⁠√(cos²θ⁠ + sin²θ + 1) = ⁠√2.⁠
cos φ = 1⁠√2,⁠ so φ = π4.

Answer: (d) 2

Method. cos²α + cos²β + cos²γ = 1 for direction angles

a⃗·b⃗ = sin²α + sin²β + sin²γ.
= 3 − (cos²α + cos²β + cos²γ) = 3 − 1 = 2.

Answer: (a) (1, ⁠√2,⁠ 1)

Method. components of a vector = magnitude × direction cosines, with l² + m² + n² = 1

Angles with 2î, 3ĵ, 4k̂ are the angles with the x-, y-, z-axes.
l = cosπ3 = 12, m = cosπ4 = 1⁠√2.⁠
n² = 1 − 14 − 12 = 14; θ acute, so n = 12.
Components = 2(l, m, n) = (1, ⁠√2,⁠ 1).

Answer: (b) 49

Method. direction cosine cos α = a₁|a⃗|

|a⃗| = ⁠√(16⁠ + 64 + 1) = ⁠√81⁠ = 9.
cos α = 49.

Answer: (a) −3281

Method. cos 2β = 2cos²β − 1 with direction cosines

cos β = −89, cos γ = 19.
cos 2β + cos 2γ = 2(cos²β + cos²γ) − 2 = 2(6481 + 181) − 2 = 13081 − 16281 = −3281.

Shortcut. cos 2β + cos 2γ = 2(1 − cos²α) − 2 = −⁠2cos²α = −⁠2·1681 = −3281.

Answer: (c) 2⁠√3⁠

Method. AB⃗ = OB⃗ − OA⃗

AB⃗ = [1 − (⁠√2⁠ − 1)]î + [(⁠√2⁠ + 1) − (−⁠1)]ĵ = (2 − ⁠√2)î⁠ + (2 + ⁠√2)ĵ.⁠
|AB⃗|² = (2 − ⁠√2)²⁠ + (2 + ⁠√2)²⁠ = (6 − 4⁠√2)⁠ + (6 + 4⁠√2)⁠ = 12.
|AB⃗| = ⁠√12⁠ = 2⁠√3.⁠

Answer: (c) ⁠√2,⁠ ⁠√2⁠

Method. equal inclination to the axes means equal direction cosines

Direction cosines of r⃗ are a|r⃗| and b|r⃗|; equal inclination to the x- and y-axes makes them equal, so a = b.
|r⃗| = 2 ⇒ a² + b² = 4 ⇒ 2a² = 4 ⇒ a = b = ⁠√2⁠ (taking the positive pair offered).
So a = ⁠√2,⁠ b = ⁠√2.⁠

Answer: (a) ⁠√3⁠ î + ĵ2

Method. unit vector in the xy-plane: â = cos θ î + sin θ ĵ

A unit vector in the xy-plane at angle θ to the positive x-axis is â = cos θ î + sin θ ĵ.
With θ = 30°: â = cos 30° î + sin 30° ĵ = ⁠√32î⁠ + 12ĵ = ⁠√3⁠ î + ĵ2. The angle is measured anticlockwise from the positive x-axis, as usual; option (b) is the vector at −⁠30° (clockwise), and (c), (d) are at ±60°.

Answer: (c) 6⁠√2⁠ î + 6ĵ ± 6k̂

Method. direction cosines: cos²α + cos²β + cos²γ = 1

cos α = cos 45° = 1⁠√2,⁠ cos β = cos 60° = 12.
cos²γ = 1 − 12 − 14 = 14, so cos γ = ±12.
OA⃗ = 12(cos α î + cos β ĵ + cos γ k̂) = 12(1⁠√2î⁠ + 12ĵ ± 12k̂)
= 6⁠√2⁠ î + 6ĵ ± 6k̂.

Answer: (d) ⁠√53⁠

Method. AB⃗ = position vector of B − position vector of A

AB⃗ = (2 − 3)î + (4 + 2)ĵ + (-3 − 1)k̂ = -î + 6ĵ − 4k̂.
|AB⃗| = ⁠√(1⁠ + 36 + 16) = ⁠√53.⁠

Answer: (b) 2

Method. scalar projection of a⃗ on b⃗ = a⃗·b⃗|b⃗|

a⃗·b⃗ = 2λ − 1 + 2λ = 4λ − 1; |b⃗| = ⁠√(5⁠ + λ²).
4λ − 1⁠√(5⁠ + λ²) = 73 ⇒ 9(4λ − 1)² = 49(5 + λ²) with 4λ − 1 > 0.
144λ² - 72λ + 9 = 245 + 49λ² ⇒ 95λ² - 72λ − 236 = 0 ⇒ (λ − 2)(95λ + 118) = 0.
λ = -11895 makes 4λ − 1 negative, so λ = 2.

Shortcut. Try the options: λ = 2 gives 7⁠√9⁠ = 73.

Answer: (d) 2π3

Method. |a⃗ + b⃗|² = |a⃗|² + |b⃗|² + 2a⃗·b⃗

|a⃗ + b⃗|² = 1 + 1 + 2a⃗·b⃗ = 1, so a⃗·b⃗ = -12.
cos θ = -12, θ = 2π3.

Answer: (c) π2

Method. expand |a⃗ + b⃗ + c⃗|² using the known dot products

As in the previous item, |a⃗ + b⃗| = 1 gives a⃗·b⃗ = -12, and |b⃗ + c⃗| = 1 gives b⃗·c⃗ = -12.
|a⃗ + b⃗ + c⃗|² = 3 + 2(a⃗·b⃗ + b⃗·c⃗ + c⃗·a⃗) = 3 + 2(-1 + c⃗·a⃗) = 1.
So c⃗·a⃗ = 0 and the angle is π2.

Answer: (a) 90°

Method. angle between diagonals from AC⃗·BD⃗

AC⃗ = (2 − 3)î + (-3 − 4)ĵ + (2 + 2)k̂ = -î − 7ĵ + 4k̂.
BD⃗ = (6 − 4)î + (-2 + 4)ĵ + (1 + 3)k̂ = 2î + 2ĵ + 4k̂.
AC⃗·BD⃗ = -2 − 14 + 16 = 0, so the diagonals are perpendicular: 90°.

Answer: (a) 0

Method. perpendicular ⇒ dot product zero; then cos 2θ = 2cos²θ − 1

(a⃗ + 2b⃗)·(5a⃗ - 4b⃗) = 5|a⃗|² - 4a⃗·b⃗ + 10a⃗·b⃗ - 8|b⃗|² = 0.
With unit vectors: 5 + 6 cos θ − 8 = 0, so cos θ = 12.
cos 2θ = 2 · 14 − 1 = -12.
cos θ + cos 2θ = 12 − 12 = 0.

Answer: (c) Both I and II

Method. test each candidate for |c⃗| = |a⃗| and c⃗·a⃗ = c⃗·b⃗ = a⃗·b⃗

|a⃗| = |b⃗| = ⁠√2⁠ and a⃗·b⃗ = 1, so c⃗ needs |c⃗| = ⁠√2,⁠ c⃗·a⃗ = 1, c⃗·b⃗ = 1.
I. î + k̂: |c⃗| = ⁠√2;⁠ c⃗·a⃗ = 1; c⃗·b⃗ = 1. Works.
II. 13(-1, 4, -1): |c⃗|² = 1 + 16 + 19 = 2; c⃗·a⃗ = -1 + 43 = 1; c⃗·b⃗ = 4 − 13 = 1. Works.
Both I and II are possible.

Answer: (b) π3

Method. cos of angle = a⃗·(a⃗ - b⃗)|a⃗||a⃗ - b⃗|

a⃗·b⃗ = 4 · 4 · cosπ3 = 8.
|a⃗ - b⃗|² = 16 + 16 − 2(8) = 16, so |a⃗ - b⃗| = 4.
a⃗·(a⃗ - b⃗) = 16 − 8 = 8.
cos φ = 84 · 4 = 12, so φ = π3.

Shortcut. a⃗, b⃗, a⃗ - b⃗ form an equilateral triangle (two sides 4 at 60°), so every angle is π3.

Answer: (b) 2⁠√17⁠

Method. projection of a⃗ on b⃗ = a⃗·b⃗|b⃗|

a⃗·b⃗ = (1)(2) + (2)(3) + (3)(−⁠2) = 2 + 6 − 6 = 2.
|b⃗| = ⁠√(4⁠ + 9 + 4) = ⁠√17.⁠
Length of projection = 2⁠√17.⁠

Answer: (c) 0 ≤ θ ≤ π2

Method. sign of a⃗·b⃗ = |a⃗||b⃗|cos θ, with 0 ≤ θ ≤ π

a⃗·b⃗ = |a⃗||b⃗|cos θ ≥ 0 with |a⃗||b⃗| > 0 ⇒ cos θ ≥ 0.
For θ in [0, π], cos θ ≥ 0 ⇔ 0 ≤ θ ≤ π2.
(θ = 0 and θ = π2 are allowed since equality is permitted, so d is too narrow.)

Answer: (c) π3

Method. perpendicular ⇒ dot product zero; |a⃗| = |b⃗| = 1

(a⃗ + 2b⃗)·(5a⃗ − 4b⃗) = 5|a⃗|² − 4a⃗·b⃗ + 10a⃗·b⃗ − 8|b⃗|² = 5 + 6a⃗·b⃗ − 8.
Set to 0: 6a⃗·b⃗ = 3 ⇒ cos θ = a⃗·b⃗ = 12.
θ = π3.

Answer: (a) 0 < x < 2

Method. obtuse angle ⇔ dot product < 0 (and vectors not opposite)

Dot product = 2x²(1) + 3x(−⁠2) + 1(x²) = 3x² − 6x = 3x(x − 2).
Obtuse ⇒ 3x(x − 2) < 0 ⇒ 0 < x < 2.
The vectors are never opposite: 2x²î + 3xĵ + k̂ = λ(î − 2ĵ + x²k̂) needs λ = 2x² ≥ 0, so the angle is never π.
Answer: 0 < x < 2.

Answer: (d) π2

Method. angle C is the angle between CA⃗ and CB⃗

CA⃗ = (ĵ + k̂) − (3ĵ + 3k̂) = −⁠2ĵ − 2k̂.
CB⃗ = (3î + ĵ + 5k̂) − (3ĵ + 3k̂) = 3î − 2ĵ + 2k̂.
CA⃗·CB⃗ = 0 + 4 − 4 = 0.
So angle C = π2.

Answer: (d) −⁠1 < sinθ < 1

Method. |a⃗ − b⃗|² = 2 − 2cos 2θ = 4sin²θ for unit vectors

|a⃗ − b⃗|² = 1 + 1 − 2cos 2θ = 4sin²θ, so |a⃗ − b⃗| = 2|sin θ|.
|a⃗ − b⃗| < 2 ⇒ |sin θ| < 1 ⇒ −⁠1 < sin θ < 1.
(Option a fails for a⃗ = b⃗: then 2θ = 0, sin θ = 0, and |a⃗ − b⃗| = 0 < 2. Options b and c fail for a⃗ ⊥ b⃗: θ = π4, sin θ = 1⁠√2.)⁠
So −⁠1 < sin θ < 1.

Answer: (d) π2

Method. solve the two linear vector equations, then test a⃗·b⃗

From 2a⃗ + b⃗ = î − 2ĵ: b⃗ = î − 2ĵ − 2a⃗.
Substitute: a⃗ + 3î − 6ĵ − 6a⃗ = 3î − ĵ ⇒ −⁠5a⃗ = 5ĵ ⇒ a⃗ = −⁠ĵ.
Then b⃗ = î − 2ĵ + 2ĵ = î.
a⃗·b⃗ = 0, so the angle is π2.

Answer: (a) 0

Method. perpendicularity gives a⃗·b⃗ = −⁠|a⃗|²

(a⃗ + b⃗)·a⃗ = 0 ⇒ |a⃗|² + a⃗·b⃗ = 0 ⇒ a⃗·b⃗ = −⁠|a⃗|².
|b⃗| = 2|a⃗| ⇒ |b⃗|² = 4|a⃗|².
(4a⃗ + b⃗)·b⃗ = 4a⃗·b⃗ + |b⃗|² = −⁠4|a⃗|² + 4|a⃗|² = 0.

Answer: (c) a⃗ must be perpendicular to b⃗.

Method. |a⃗ + b⃗|² − |a⃗ − b⃗|² = 4a⃗·b⃗

|a⃗ + b⃗|² = |a⃗|² + |b⃗|² + 2a⃗·b⃗ and |a⃗ − b⃗|² = |a⃗|² + |b⃗|² − 2a⃗·b⃗.
Equal magnitudes ⇒ 4a⃗·b⃗ = 0 ⇒ a⃗·b⃗ = 0, so a⃗ ⊥ b⃗.
The other options fail: e.g. a⃗ = 2⁠√2î,⁠ b⃗ = 2⁠√2ĵ⁠ satisfy the data (|a⃗ + b⃗| = |a⃗ − b⃗| = 4) without being unit, parallel or equal vectors.

Shortcut. Geometrically, equal diagonals make the parallelogram on a⃗, b⃗ a rectangle.

Answer: (c) 31

Method. diagonals a⃗ ± b⃗; (a⃗ + b⃗)·(a⃗ − b⃗) = |a⃗|² − |b⃗|²

Let a⃗ = 2î − 4ĵ + 5k̂, b⃗ = î − 2ĵ − 3k̂; the diagonals are a⃗ + b⃗ and a⃗ − b⃗.
(a⃗ + b⃗)·(a⃗ − b⃗) = |a⃗|² − |b⃗|².
|a⃗|² = 4 + 16 + 25 = 45, |b⃗|² = 1 + 4 + 9 = 14.
Dot product = 45 − 14 = 31, magnitude 31.

Answer: (b) 199

Method. scalar projection of a⃗ on b⃗ = a⃗ · b⃗|b⃗|

a⃗ · b⃗ = 4 + 8 + 7 = 19.
|b⃗| = ⁠√(16⁠ + 16 + 49) = 9.
Scalar projection = 199.

Answer: (b) The vectors are perpendicular

Method. |a⃗ + b⃗|² - |a⃗ - b⃗|² = 4a⃗ · b⃗

|a⃗ + b⃗| = |a⃗ - b⃗| ⇒ |a⃗|² + |b⃗|² + 2a⃗ · b⃗ = |a⃗|² + |b⃗|² - 2a⃗ · b⃗.
So 4a⃗ · b⃗ = 0 ⇒ a⃗ · b⃗ = 0; non-zero vectors ⇒ perpendicular.

Answer: (b) |a̅ − b̅|²4

Method. |â - b̂|² = 2 − 2 cos θ = 4 sin²(θ/⁠2) for unit vectors

|a⃗ - b⃗|² = |a⃗|² + |b⃗|² - 2a⃗ · b⃗ = 2 − 2 cos θ.
2 − 2 cos θ = 4 sin²(θ2).
So sin²(θ2) = |a⃗ - b⃗|²4 (the options' a̅, b̅ are the same vectors).

Answer: (b) -106

Method. compute the two vectors, then the dot product

b⃗ - a⃗ = î + 3ĵ − 8k̂.
3a⃗ + b⃗ = (3 + 2)î + (-6 + 1)ĵ + (15 − 3)k̂ = 5î − 5ĵ + 12k̂.
Dot product = 5 − 15 − 96 = -106.

Answer: (a) p²

Method. XY⃗ · XZ⃗ = XY · XZ cos X at each vertex; right angle at B

Hypotenuse AC ⇒ ∠B = 90°.
AB⃗ · AC⃗ = AB · AC cos A = AB · AC · ABAC = AB².
BC⃗ · BA⃗ = BC · BA cos 90° = 0.
CA⃗ · CB⃗ = CA · CB cos C = CB².
Sum = AB² + BC² = AC² = p².

Answer: (d) 4

Method. perpendicular vectors: dot product is zero

3(-2) + 4λ + (-1)(10) = 0 ⇒ 4λ − 16 = 0 ⇒ λ = 4.

Answer: (a) a⃗ and b⃗ are perpendicular

Method. expand the dot product

(a⃗ + b⃗) · (a⃗ + b⃗) = |a⃗|² + 2a⃗ · b⃗ + |b⃗|².
This equals |a⃗|² + |b⃗|² ⇔ a⃗ · b⃗ = 0 ⇔ a⃗ ⊥ b⃗ (both non-zero).

Answer: (d) (x + y + z)

Method. dot product of components

r⃗ · (î + ĵ + k̂) = x(1) + y(1) + z(1) = x + y + z.

Answer: (d) 5

Method. |a⃗ ± b⃗|² = |a⃗|² + |b⃗|² ± 2a⃗ · b⃗

25 = 9 + 16 − 2a⃗ · b⃗ ⇒ a⃗ · b⃗ = 0.
|a⃗ + b⃗|² = 9 + 16 + 0 = 25 ⇒ |a⃗ + b⃗| = 5.

Shortcut. |a⃗ + b⃗|² + |a⃗ - b⃗|² = 2(|a⃗|² + |b⃗|²) = 50, so |a⃗ + b⃗|² = 25.

Answer: (c) |A⃗| = |B⃗| = |C⃗|

Method. magnitude via an orthonormal set (all cross dot products vanish)

For mutually perpendicular unit vectors, |xa⃗ + yb⃗ + zc⃗|² = x² + y² + z².
|A⃗|² = 1 + 1 + 1 = 3, |B⃗|² = 1 + 1 + 1 = 3, |C⃗|² = 1 + 1 + 1 = 3.
So |A⃗| = |B⃗| = |C⃗| = ⁠√3.⁠

Answer: (a) 9p² = 4q²

Method. section formula (internal and external), then dot product zero

R divides PQ internally in 2 : 3: OR⃗ = 3p⃗ + 2q⃗5.
S divides PQ externally in 2 : 3: OS⃗ = 2q⃗ - 3p⃗2 − 3 = 3p⃗ - 2q⃗.
OR⃗ · OS⃗ = 0 ⇒ (3p⃗ + 2q⃗) · (3p⃗ - 2q⃗) = 9p² - 4q² = 0.
So 9p² = 4q² (p, q the magnitudes).

Answer: (c) 9 units

Method. work done = F⃗ · displacement

Displacement AB⃗ = (3 − 1)î + (-1 − 2)ĵ + (5 + 3)k̂ = 2î − 3ĵ + 8k̂.
W = F⃗ · AB⃗ = 1(2) + 3(-3) + 2(8) = 2 − 9 + 16 = 9 units.

No correct option. This question is defective as printed in UPSC's paper.

Method. perpendicular vectors: dot product is zero

a⃗ · b⃗ = 2(3) + 3(2) + 4λ = 12 + 4λ = 0 ⇒ λ = -3.
-3 is not among the options 2, 3, 4, 5.

Perpendicularity forces λ = -3, which no option offers; if the minus sign was lost in printing, the intended answer is (b) 3.

Answer: (d) 4 square units

Method. area = |(a⃗ + 3b⃗) × (3a⃗ + b⃗)|

(a⃗ + 3b⃗) × (3a⃗ + b⃗) = a⃗ × b⃗ + 9(b⃗ × a⃗) = -8(a⃗ × b⃗).
|a⃗ × b⃗| = 1 · 1 · sin 30° = 12.
Area = 8 · 12 = 4 square units.

Answer: (c) 2(b⃗ × c⃗)

Method. a⃗ ⊥ b⃗, c⃗ ⇒ a⃗ is the unit vector along ±b⃗ × c⃗

a⃗·b⃗ = a⃗·c⃗ = 0, so a⃗ is parallel to b⃗ × c⃗.
|b⃗ × c⃗| = 1 · 1 · sinπ6 = 12.
a⃗ = ±b⃗ × c⃗1/⁠2 = ±2(b⃗ × c⃗); among the options, 2(b⃗ × c⃗).

Answer: (d) 4⁠√3⁠

Method. (a⃗ - b⃗) × (a⃗ + b⃗) = 2(a⃗ × b⃗) and Lagrange's identity

p⃗ × q⃗ = a⃗ × a⃗ + a⃗ × b⃗ - b⃗ × a⃗ - b⃗ × b⃗ = 2(a⃗ × b⃗).
|a⃗ × b⃗|² = |a⃗|²|b⃗|² - (a⃗·b⃗)² = 16 − 4 = 12, so |a⃗ × b⃗| = 2⁠√3.⁠
|p⃗ × q⃗| = 2 · 2⁠√3⁠ = 4⁠√3.⁠

Answer: (d) All three

Method. every vector perpendicular to both is a non-zero scalar multiple of their cross product

(2î − ĵ + k̂) × (î + ĵ + 3k̂) = (-3 − 1)î − (6 − 1)ĵ + (2 + 1)k̂ = -4î − 5ĵ + 3k̂.
I. 4î + 5ĵ − 3k̂ = -1 × (cross product).
II. -8î − 10ĵ + 6k̂ = 2 × (cross product).
III. 150(-4î − 5ĵ + 3k̂) = 150 × (cross product).
All three are perpendicular to both vectors.

Answer: (c) ⁠√26⁠ square units

Method. area of parallelogram = |a⃗ × b⃗|

(î + 2ĵ + 3k̂) × (2î + ĵ + 2k̂) = (4 − 3)î − (2 − 6)ĵ + (1 − 4)k̂ = î + 4ĵ − 3k̂.
Area = ⁠√(1⁠ + 16 + 9) = ⁠√26⁠ square units.

Answer: (a) -2

Method. moment about B = BA⃗ × F⃗; compare components

BA⃗ = (1 + 1)î + (2 + 2)ĵ + (5 − 3)k̂ = 2î + 4ĵ + 2k̂.
BA⃗ × F⃗ = (4.5 − 2(-λ))î − (2.5 − 2.2)ĵ + (2(-λ) − 4.2)k̂ = (20 + 2λ)î − 6ĵ − (2λ + 8)k̂.
î: 20 + 2λ = 16 gives λ = -2.
k̂ check: -(2λ + 8) = 2λ gives λ = -2 also.

Answer: (a) 0⃗

Method. substitute c⃗ and use a⃗ × a⃗ = b⃗ × b⃗ = 0⃗

b⃗ × c⃗ = cos²θ(b⃗ × a⃗) + sin²θ(b⃗ × b⃗) = cos²θ(b⃗ × a⃗).
c⃗ × a⃗ = cos²θ(a⃗ × a⃗) + sin²θ(b⃗ × a⃗) = sin²θ(b⃗ × a⃗).
Sum = a⃗ × b⃗ + (cos²θ + sin²θ)(b⃗ × a⃗) = a⃗ × b⃗ - a⃗ × b⃗ = 0⃗.

Shortcut. Coefficients cos²θ + sin²θ = 1, so C lies on AB; the expression is (b⃗ - a⃗) × (c⃗ - a⃗) = 0⃗ for collinear points.

Answer: (a) 0

Method. |a⃗ × b⃗| = |a⃗||b⃗| sin θ

|a⃗ × b⃗| = 1 · 1 · sin θ = 1, so sin θ = 1 and θ = π2.
a⃗·b⃗ = cosπ2 = 0.

Shortcut. |a⃗ × b⃗|² + (a⃗·b⃗)² = |a⃗|²|b⃗|² gives 1 + (a⃗·b⃗)² = 1.

Answer: (a) 36

Method. cross product is null ⇔ the vectors are parallel ⇔ components proportional

(2î + 6ĵ + 27k̂) × (î + αĵ + βk̂) = 0⃗ means the two vectors are parallel.
12 = α6 = β27, so α = 3, β = 272.
3α + 2β = 9 + 27 = 36.

Answer: (b) 30°

Method. write the expression as |a⃗||b⃗|(sin θ + ⁠√3|cos⁠ θ|) and use R sin(θ + α) form

|a⃗ × b⃗| + ⁠√3|a⃗·b⃗|⁠ = |a⃗||b⃗|(sin θ + ⁠√3|cos⁠ θ|), 0 ≤ θ ≤ π.
The bracket is symmetric about θ = 90°, so take 0 ≤ θ ≤ 90°: sin θ + ⁠√3⁠ cos θ = 2 sin(θ + 60°).
Maximum value 2 when θ + 60° = 90°, i.e. θ = 30°.
Check: at 0° and 60° the bracket is ⁠√3,⁠ at 45° it is about 1.93, at 30° it is 2.

Answer: (d) Neither 1 nor 2

Method. moment about a point is r⃗ × F⃗; moment about a line is the scalar projection of that on the line

Moment of F⃗ about O is M⃗ = r⃗ × F⃗, where r⃗ is the position vector (from O) of the point of application.
Changing the point of application to one off the line of action changes r⃗ × F⃗ (it is unchanged only for points on the same line of action), so Statement 1 is false.
Moment about a line through O with unit vector n̂ is (r⃗ × F⃗) · n̂, a scalar, so Statement 2 is false.

Answer: (a) 0⃗

Method. take r⃗ × out as a common factor and use r⃗ = (r⃗·î)î + (r⃗·ĵ)ĵ + (r⃗·k̂)k̂

Each term is (r⃗·î)(r⃗ × î) = r⃗ × ((r⃗·î)î), since (r⃗·î) is a scalar.
Sum = r⃗ × [(r⃗·î)î + (r⃗·ĵ)ĵ + (r⃗·k̂)k̂] = r⃗ × r⃗.
r⃗ × r⃗ = 0⃗.

Answer: (a) 3

Method. Lagrange identity: (a⃗ × b⃗)² + (a⃗ · b⃗)² = |a⃗|²|b⃗|²

(a⃗ × b⃗)² + (a⃗ · b⃗)² = |a⃗|²|b⃗|²(sin²θ + cos²θ) = |a⃗|²|b⃗|².
16|a⃗|² = 144 ⇒ |a⃗|² = 9 ⇒ |a⃗| = 3.

Answer: (b) 2 only

Method. unit vectors ⊥ to two non-parallel vectors come in a ± pair; angle from the dot product

a⃗ × b⃗ = (1, 1, -1) ≠ 0⃗, so a⃗ and b⃗ are not parallel.
The unit vectors perpendicular to both are ±a⃗ × b⃗|a⃗ × b⃗| = ±1⁠√3(1,⁠ 1, -1): two of them, so Statement 1 is false.
a⃗ · b⃗ = 0 + 0 + 1 = 1, |a⃗| = |b⃗| = ⁠√2.⁠
cos θ = 1⁠√2⁠ · ⁠√2⁠ = 12, so θ = π3: Statement 2 is true.

Answer: (b) 5î + 2ĵ + 2k̂

Method. test the options against a⃗·b⃗ = 27 and a⃗ × b⃗ = 9c⃗

Let b⃗ = xî + yĵ + zk̂. a⃗·b⃗ = 3(x + y + z) = 27 ⇒ x + y + z = 9 (rules out d).
a⃗ × b⃗ = 3[(z − y)î + (x − z)ĵ + (y − x)k̂] = 9ĵ − 9k̂ ⇒ z = y, x − z = 3.
With x + y + z = 9: x + 2y = 9 and x = y + 3 ⇒ y = 2, z = 2, x = 5.
b⃗ = 5î + 2ĵ + 2k̂.

Answer: (a) π2

Method. angle via dot product

From the previous item, b⃗ = 5î + 2ĵ + 2k̂, so a⃗ + b⃗ = 8î + 5ĵ + 5k̂.
(a⃗ + b⃗)·c⃗ = 0 + 5 − 5 = 0.
The angle is π2.

Shortcut. c⃗ is parallel to a⃗ × b⃗, hence perpendicular to both a⃗ and b⃗, so to a⃗ + b⃗.

Answer: (c) 1 and 2 only

Method. algebraic laws of dot and cross products

1. a⃗·(b⃗ + c⃗) = a⃗·b⃗ + a⃗·c⃗. Correct.
2. a⃗ × (b⃗ + c⃗) = a⃗ × b⃗ + a⃗ × c⃗. Correct.
3. Not associative: (î × î) × ĵ = 0⃗, but î × (î × ĵ) = î × k̂ = −⁠ĵ. Not correct.
So 1 and 2 only.

Answer: (b) 2 only

Method. a⃗ × b⃗ is determined by a⃗, b⃗; but (a⃗ + λb⃗) × b⃗ = a⃗ × b⃗

2. Given a⃗ and b⃗, the cross product a⃗ × b⃗ is a single definite vector, so c⃗ is unique. Correct.
1. For any scalar λ, (a⃗ + λb⃗) × b⃗ = a⃗ × b⃗ + λ(b⃗ × b⃗) = c⃗. So infinitely many vectors give the same c⃗; a⃗ is not unique. Not correct.
Only 2 is correct.

Answer: (b) 2 only

Method. a⃗·(b⃗ × c⃗) = c⃗·(a⃗ × b⃗); a⃗ × b⃗ is normal to the plane of a⃗, b⃗

2. a⃗·(b⃗ × c⃗) = [a⃗ b⃗ c⃗] = (a⃗ × b⃗)·c⃗ = 0, since a⃗ × b⃗ ⊥ c⃗. Correct.
1. a⃗ × b⃗ = sin θ n̂, where n̂ is the unit normal to a⃗ and b⃗. But c⃗ ⊥ (a⃗ × b⃗), so c⃗ cannot be along n̂; sin θ c⃗ is a vector perpendicular to a⃗ × b⃗ and cannot equal it unless sin θ = 0. Not correct in general.
Only 2 is correct.

Answer: (c) Both 1 and 2

Method. (a⃗ + b⃗)·(a⃗ − b⃗) = |a⃗|² − |b⃗|²; a⃗ × b⃗ is perpendicular to both a⃗ and b⃗

1. c⃗·(a⃗ − b⃗) = (a⃗ + b⃗)·(a⃗ − b⃗) = |a⃗|² − |b⃗|² = 0, since |a⃗| = |b⃗|. Correct.
2. a⃗ × b⃗ is perpendicular to a⃗ and to b⃗, so c⃗·(a⃗ × b⃗) = a⃗·(a⃗ × b⃗) + b⃗·(a⃗ × b⃗) = 0 + 0 = 0. Correct.
Both statements are correct.

Answer: (d) 1, 2 and 3

Method. |â × b̂| = sin θ, â·b̂ = cos θ, |â ± b̂|² = 2 ± 2cos θ

For unit vectors at angle θ:
1. |â × b̂| = sin θ, which is 1 only when θ = π2 (e.g. î × î = 0⃗). Not correct.
2. â·b̂ = cos θ, which is 1 only when θ = 0 (e.g. î·ĵ = 0). Not correct.
3. |â + b̂|² = 2 + 2cos θ and |â − b̂|² = 2 − 2cos θ; the sum is larger only when θ < π2 (for î, ĵ they are equal). Not correct.
All three are not correct.

Answer: (a) 3

Method. Lagrange identity: |a⃗ × b⃗|² + (a⃗ · b⃗)² = |a⃗|²|b⃗|²

|a⃗ × b⃗|² + (a⃗ · b⃗)² = |a⃗|²|b⃗|²(sin²θ + cos²θ) = |a⃗|²|b⃗|².
So 16|b⃗|² = 144, |b⃗|² = 9, |b⃗| = 3.

Answer: (c) 1 and 3 only

Method. test each identity; Lagrange's identity for 3

1. (a⃗ + b⃗) · (a⃗ - b⃗) = |a⃗|² - a⃗ · b⃗ + b⃗ · a⃗ - |b⃗|² = |a⃗|² - |b⃗|²: correct.
2. Counter-example: a⃗ = î, b⃗ = ĵ: LHS = ⁠√2⁠ · ⁠√2⁠ = 2, RHS = 1 − 1 = 0: wrong.
3. (a⃗ · b⃗)² + |a⃗ × b⃗|² = |a⃗|²|b⃗|²(cos² θ + sin² θ) = |a⃗|²|b⃗|²: correct.
So 1 and 3 only.

Answer: (b) 2 only

Method. |a⃗ × b⃗| is the area of the parallelogram, twice the triangle

1. |a⃗ × b⃗| = |a⃗||b⃗| sin θ is the area of the parallelogram on a⃗, b⃗; the triangle has half of it: wrong.
2. a⃗ × b⃗ = 0⃗ with both non-zero ⇒ sin θ = 0 ⇒ a⃗ ∥ b⃗ ⇒ a⃗ = λb⃗: correct.
So 2 only.

Answer: (b) 5⁠√26⁠

Method. cos θ from the dot product, then sin θ = ⁠√(1⁠ −⁠ cos² θ)

a⃗ · b⃗ = 8 − 18 + 3 = -7; |a⃗| = ⁠√(4⁠ + 36 + 9) = 7; |b⃗| = ⁠√(16⁠ + 9 + 1) = ⁠√26.⁠
cos θ = -77⁠√26⁠ = -1⁠√26.⁠
sin θ = ⁠√(1⁠ −⁠ ⁠126⁠) = 5⁠√26.⁠

Answer: (a) 1⁠√3⁠ î + 1⁠√3⁠ ĵ − 1⁠√3⁠ k̂

Method. unit vector along the cross product

 î ĵ k̂ 2 -1 1 3 -4 -1  = î(1 + 4) − ĵ(-2 − 3) + k̂(-8 + 3) = 5î + 5ĵ − 5k̂.
Unit vector: 1⁠√3(î⁠ + ĵ − k̂), which is option (a).
(c) and (d) fail: their dot products with 2î − ĵ + k̂ are 2⁠√3⁠ ≠ 0; (b) is a unit vector but is not perpendicular either: its dot product with 2î − ĵ + k̂ is ⁠√2⁠ ≠ 0.

Answer: (c) 2(a⃗ × b⃗)

Method. distribute the cross product; a⃗ × a⃗ = 0⃗

(a⃗ - b⃗) × (a⃗ + b⃗) = a⃗ × a⃗ + a⃗ × b⃗ - b⃗ × a⃗ - b⃗ × b⃗.
= 0⃗ + a⃗ × b⃗ + a⃗ × b⃗ - 0⃗ = 2(a⃗ × b⃗).

Answer: (c) ⁠√5⁠ λ

Method. moment about the origin = r⃗ × F⃗

r⃗ × F⃗ =  î ĵ k̂ 1 2 3 0 0 λ  = 2λî − λĵ + 0k̂.
Magnitude = ⁠√(4λ²⁠ + λ²) = ⁠√5⁠ λ (λ > 0).

Answer: (a) 30°

Method. |a⃗ × b⃗| = |a⃗||b⃗| sin θ

|a⃗ × b⃗| = ⁠√(9⁠ + 4 + 36) = 7.
sin θ = 72 × 7 = 12 ⇒ acute θ = 30°.

Answer: (a) -3î + 11ĵ + 9k̂

Method. moment = r⃗ × F⃗, r⃗ drawn from the point to the line of action

r⃗ = (2î − ĵ + 3k̂) − (î + 2ĵ − k̂) = î − 3ĵ + 4k̂.
r⃗ × F⃗ =  î ĵ k̂ 1 -3 4 3 0 1  = î(-3 − 0) − ĵ(1 − 12) + k̂(0 + 9).
= -3î + 11ĵ + 9k̂.

Answer: (d) 6

Method. eliminate a⃗ using the given relation

a⃗ = -2b⃗ - 3c⃗.
a⃗ × b⃗ = -3c⃗ × b⃗ = 3(b⃗ × c⃗).
c⃗ × a⃗ = -2c⃗ × b⃗ = 2(b⃗ × c⃗).
Sum = (3 + 1 + 2)(b⃗ × c⃗) = 6(b⃗ × c⃗), so λ = 6.

Answer: (b) 0⃗

Method. cross product of parallel vectors is the zero vector

k⃗ ∥ A⃗ ⇒ k⃗ × A⃗ = 0⃗.
So kk⃗ × A⃗ = k(0⃗) = 0⃗.
The evidently intended k⃗ × (k⃗ × A⃗) = k⃗ × 0⃗ = 0⃗ as well, so (b) either way.

Answer: (b) 2⁠√3⁠ units

Method. a vector perpendicular to two vectors is parallel to their cross product

α⃗ × β⃗ =  î ĵ k̂ 1 2 -1 2 -1 3  = 5î − 5ĵ − 5k̂.
So δ⃗ = t(î − ĵ − k̂).
δ⃗ · γ⃗ = t(2 − 1 − 6) = -5t = 10 ⇒ t = -2.
|δ⃗| = 2⁠√3⁠ units.

Answer: (b) 2|a⃗|²

Method. |a⃗ × û|² = |a⃗|² - (a⃗ · û)² for a unit vector

Let a⃗ = a₁î + a₂ĵ + a₃k̂.
a⃗ × î = a₃ĵ − a₂k̂, so |a⃗ × î|² = a₂² + a₃².
Likewise |a⃗ × ĵ|² = a₁² + a₃², |a⃗ × k̂|² = a₁² + a₂².
Sum = 2(a₁² + a₂² + a₃²) = 2|a⃗|².

Answer: (d) 0

Method. every scalar triple product of coplanar vectors is zero

(3a⃗ + 2b⃗) and (5a⃗ − 4c⃗) lie in the plane of a⃗, b⃗, c⃗, so their cross product is normal to that plane (or zero).
(b⃗ + 2c⃗) lies in the same plane, so the dot product of that normal vector with it is 0.
Term by term: the expansion is a sum of multiples of [a⃗ c⃗ b⃗], [b⃗ a⃗ c⃗], [b⃗ c⃗ b⃗], … each zero for coplanar vectors.
Value = 0.

Answer: (c) 0

Method. scalar triple product of coplanar vectors is zero

(2a⃗ × 3b⃗)·4c⃗ = 24[a⃗ b⃗ c⃗] and (5b⃗ × 3c⃗)·6a⃗ = 90[b⃗ c⃗ a⃗] = 90[a⃗ b⃗ c⃗] (cyclic order).
Sum = 114[a⃗ b⃗ c⃗].
Coplanar ⇒ [a⃗ b⃗ c⃗] = 0, so the sum is 0.

Answer: (b) -1

Method. coplanar vectors: scalar triple product (determinant of components) = 0

Coplanar ⇔  2 -3 1 1 2 -3 0 1 p  = 0 (determinant).
Expand along row 1: 2(2p + 3) + 3(p − 0) + 1(1 − 0) = 4p + 6 + 3p + 1 = 7p + 7.
7p + 7 = 0 ⇒ p = −⁠1.

Answer: (b) 1

Method. coplanarity determinant, simplified by row operations

 a 1 1 1 b 1 1 1 c  = 0. Apply R₂ → R₂ - R₁, R₃ → R₃ - R₁: a 1 1 1 − a b − 1 0 1 − a 0 c − 1  = 0.
Put x = 1 − a, y = 1 − b, z = 1 − c (all non-zero). Expanding along R₁: (1 − x)yz + xz + xy = 0.
So xy + yz + zx = xyz; dividing by xyz: 1x + 1y + 1z = 1.
Hence 11 − a + 11 − b + 11 − c = 1.

Shortcut. Test a = 2, b = 3: coplanarity gives c = 3/⁠5, and -1 − 12 + 52 = 1.

Answer: (d) m = ±2 and n = 1

Method. coplanar vectors: scalar triple product is zero

Coplanar ⇒ [a⃗ b⃗ c⃗] = 0: 1 -1 1 2 3 2 1 m n  = 1(3n − 2m) + 1(2n − 2) + 1(2m − 3) = 5n − 5 = 0, so n = 1.
|c⃗|² = 1 + m² + n² = 6 ⇒ m² = 4 ⇒ m = ±2.
So m = ±2 and n = 1.

Answer: (c) Three

Method. dot and cross products are defined only between two vectors

I. a⃗ × b⃗ and c⃗ × d⃗ are vectors; their dot product is defined.
II. Cross product of two vectors: defined.
III. a⃗·b⃗ and c⃗·d⃗ are scalars; a dot product of two scalars is not defined.
IV. Cross product of two scalars: not defined.
V. (a⃗ × b⃗)·c⃗ is a scalar; its cross product with d⃗ is not defined.
Three (III, IV, V) are not well-defined.

Answer: (a) I only

Method. a cross product is perpendicular to both factors; compare magnitudes

c⃗ = a⃗ × b⃗ gives c⃗ ⊥ a⃗, c⃗ ⊥ b⃗; a⃗ = b⃗ × c⃗ gives a⃗ ⊥ b⃗. So I is true.
With the right angles, |c⃗| = |a⃗||b⃗| and |a⃗| = |b⃗||c⃗|.
So |c⃗| = |b⃗|²|c⃗|, giving |b⃗| = 1 and |c⃗| = |a⃗|, but |a⃗| is free.
Example: a⃗ = 2î, b⃗ = ĵ, c⃗ = 2k̂ satisfies both relations, and a⃗, c⃗ are not unit. So II is not necessarily true.

Answer: (d) Neither I nor II

Method. substitute the given relations; each scalar triple product equals |c⃗|² = |a⃗|²

(a⃗ × b⃗)·c⃗ = c⃗·c⃗ = |c⃗|², (b⃗ × c⃗)·a⃗ = |a⃗|², (c⃗ × a⃗)·b⃗ = [a⃗ b⃗ c⃗] = |c⃗|².
I reads |c⃗|² + |a⃗|² = |c⃗|², i.e. |a⃗| = 0, impossible for non-zero vectors: I is false.
(a⃗ × b⃗) × (b⃗ × c⃗) = c⃗ × a⃗, so II's left side is (c⃗ × a⃗)·b⃗ = |c⃗|².
From the previous item |c⃗| = |a⃗| is not fixed (e.g. a⃗ = 2î, b⃗ = ĵ, c⃗ = 2k̂ gives 4), so II is not necessarily true.
Neither statement holds in general.