Understanding Scalars and Vectors: A Comprehensive Guide

Vector Addition (Resultant Bisecting Angle Condition) JEE Main 2024 (Jan Shift 2)

⏱ JEE Standard Pace: 120s +4 / -1 Scale

Q1. Two vectors A and B are inclined at an angle θ = 120°. The resultant of A and B makes an angle of α₁ with vector A, while the difference vector (A − B) makes an angle of α₂ with vector A. If |A| = |B|, what are the values of α₁ and α₂ respectively?

A 60° and 60° B 30° and 60° C 60° and 30° D 45° and 45°

✔ Rhombus Diagonal Bisection (θ/2 = 60° and (180° − θ)/2 = 30°) Mastered! +4 Marks

✖ Since |A| = |B|, A + B bisects the 120° angle between A and B (α₁ = 60°), while A − B bisects the 60° angle between A and (−B) (α₂ = 30°)!

Concept & Solution:
• When |A| = |B|, the parallelogram of vectors is a rhombus whose diagonals bisect the vertex angles:
• Angle of (A + B) with A: α₁ = θ / 2 = 120° / 2 = 60°.
• Angle between A and (−B) is 180° − 120° = 60°, so angle of (A − B) with A is α₂ = 60° / 2 = 30° (also α₁ + α₂ = 90° because diagonals of a rhombus are perpendicular!).

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Lami’s Theorem (Two Smooth Inclined Planes Wedge) JEE Main 2025 (Jan Shift 1)

⏱ JEE Standard Pace: 120s+4 / -1 Scale

Q2. A uniform smooth cylinder of weight W = 100 N rests in static equilibrium in a V-shaped groove formed by two frictionless inclined planes making angles of 30° and 60° with the horizontal respectively, as shown below. What are the normal reactions N₁ (from the 30° plane) and N₂ (from the 60° plane) on the cylinder?

A N₁ = 50 N ; N₂ = 50√3 N B N₁ = 50√3 N ; N₂ = 50 N C N₁ = 100 / √3 N ; N₂ = 200 / √3 N D N₁ = 50√2 N ; N₂ = 50√2 N

✔ Lami’s Theorem on Normal Forces (N₁ Makes 30° with Vertical) Mastered! +4 Marks

✖ Normal to a plane inclined at θ to the horizontal makes angle θ with the VERTICAL! Thus N₁ makes 30° with vertical (N₁ = W cos 30° = 50√3 N) and N₂ makes 60° with vertical (N₂ = W cos 60° = 50 N)!

Concept & Solution:
• The normal N₁ to the 30° plane makes an angle of 30° with the upward vertical.
• The normal N₂ to the 60° plane makes an angle of 60° with the upward vertical.
• Since 30° + 60° = 90°, N₁ and N₂ are perpendicular to each other.
• By Lami’s Theorem: N₁ = W cos 30° = 50√3 N, and N₂ = W cos 60° = 50 N.

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Resolution of Vectors (Rotated Coordinate Axes X’-Y’) JEE Main 2024 (Apr Shift 1)

⏱ JEE Standard Pace: 120s+4 / -1 Scale

Q3. In the standard x-y coordinate system, a vector A has components A_x = 2√3 and A_y = 2. If the coordinate axes (x and y) are rotated counter-clockwise in the same plane through an angle of 30° (keeping vector A fixed in space), what will be the new components (A_x’, A_y’) of vector A along the rotated axes?

A A_x’ = 4 ; A_y’ = 0 B A_x’ = 0 ; A_y’ = 4 C A_x’ = 2 ; A_y’ = 2√3 D A_x’ = 2√2 ; A_y’ = 2√2

✔ Polar Angle Insight (θ_A = 30° Aligns Exactly with New X’-Axis!) +4 Marks

✖ Magnitude |A| = √((2√3)² + 2²) = 4, and tan θ = 2/(2√3) = 1/√3 ⇒ θ = 30°! Rotating the x-axis by +30° aligns the new x’-axis directly along A!

Concept & Solution:
• Magnitude of A: |A| = √( (2√3)² + 2² ) = √(12 + 4) = 4.
• Angle of A with the original positive x-axis: tan θ = A_y / A_x = 2 / (2√3) = 1 / √3 ⇒ θ = 30°.
• When the axes are rotated counter-clockwise by 30°, the new x’-axis points at 30°—which is directly along vector A!
• Therefore, A_x’ = |A| cos 0° = 4 and A_y’ = |A| sin 0° = 0.

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Cross Product Condition (A × B = 0 & B × C = 0) JEE Main 2023 (Jan Shift 2)

⏱ JEE Standard Pace: 120s+4 / -1 Scale

Q4. Let A, B, and C be three non-zero vectors such that no two of them are collinear. If (A × B) = C and (B × C) = A, then what is the angle between A and B, and what must be the magnitude of vector B?

A 60° ; |B| = 2 / √3 B 90° ; |B| = 2 C 45° ; |B| = √2 D 90° ; |B| = 1 (and |A| = |C|)

✔ Mutual Orthogonality of Cross Products (|B| = 1, |A| = |C|) Mastered! +4 Marks

✖ Since A × B = C (so C ⊥ A, B) and B × C = A (so A ⊥ B, C), A, B, C are mutually perpendicular (90°)! Then |A||B| = |C| and |B||C| = |A| ⇒ |B|² = 1 ⇒ |B| = 1!

Concept & Solution:
• From A × B = C, C is perpendicular to both A and B.
• From B × C = A, A is perpendicular to both B and C.
• Thus A, B, and C form a mutually orthogonal triad (90° between each pair).
• Taking magnitudes (with sin 90° = 1): |A| |B| = |C| and |B| |C| = |A|.
• Multiplying the two equations gives |A| |C| |B|² = |A| |C| ⇒ |B| = 1 and |A| = |C|.

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Dot Product (Angle of Vector with Coordinate Plane) JEE Main 2024 (Jan Shift 1)

⏱ JEE Standard Pace: 120s+4 / -1 Scale

Q5. A vector in 3D space is given by A = (3 î + 4 ĵ + 5 k̂). What angle does vector A make with the z-axis, and what angle does it make with the x-y plane?

A 30° with z-axis ; 60° with x-y plane B 45° with z-axis ; 45° with x-y plane C 60° with z-axis ; 30° with x-y plane D cos⁻¹(5 / 12) with z-axis ; sin⁻¹(5 / 12) with x-y plane

✔ Cylindrical Projection A_xy = √(3² + 4²) = 5 = A_z Mastered! +4 Marks

✖ Magnitude in x-y plane is A_xy = √(3² + 4²) = 5, which equals A_z = 5! So tan θ = 5/5 = 1 ⇒ 45° with both the z-axis and the x-y plane!

Concept & Solution:
• Component of A in the x-y plane: A_xy = √(A_x² + A_y²) = √(3² + 4²) = 5.
• Component along the z-axis: A_z = 5; Total magnitude: |A| = √(5² + 5²) = 5√2.
• Angle with z-axis: cos γ = A_z / |A| = 5 / (5√2) = 1 / √2 ⇒ γ = 45°.
• Angle with x-y plane: 90° − γ = 90° − 45° = 45°.

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Vector Addition (Three Unit Vectors Sum to Zero) JEE Main 2025 (Apr Shift 2)

⏱ JEE Standard Pace: 120s+4 / -1 Scale

Q6. If â, b̂, and ĉ are three unit vectors such that â + b̂ + ĉ = 0, then what is the value of the scalar quantity (â · b̂ + b̂ · ĉ + ĉ · â)?

A 0 B −3 C −3 / 2 D +3 / 2

✔ Identity |â + b̂ + ĉ|² = 3 + 2(â·b̂ + b̂·ĉ + ĉ·â) = 0 Mastered! +4 Marks

✖ Square both sides of â + b̂ + ĉ = 0: |â|² + |b̂|² + |ĉ|² + 2(â·b̂ + b̂·ĉ + ĉ·â) = 0 ⇒ 3 + 2S = 0 ⇒ S = −3/2!

Concept & Solution:
• Squaring the vector equation â + b̂ + ĉ = 0 (taking dot product with itself):
• |â + b̂ + ĉ|² = |â|² + |b̂|² + |ĉ|² + 2 (â · b̂ + b̂ · ĉ + ĉ · â) = 0.
• Since |â| = |b̂| = |ĉ| = 1, we have: 1 + 1 + 1 + 2 (â · b̂ + b̂ · ĉ + ĉ · â) = 0.
• Therefore, â · b̂ + b̂ · ĉ + ĉ · â = −3 / 2.

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Scalars vs. Vectors (Polar vs. Axial Vectors / Pseudovectors) JEE Main 2022 (June Shift 1)

⏱ JEE Standard Pace: 120s+4 / -1 Scale

Q7. Which of the following sets consists exclusively of axial vectors (pseudovectors) whose direction is assigned along the axis of rotation by the right-hand screw rule?

A Angular velocity (ω), Torque (τ), and Angular momentum (L) B Linear momentum (p), Force (F), and Torque (τ) C Displacement (r), Velocity (v), and Angular acceleration (α) D Electric field (E), Force (F), and Linear acceleration (a)

✔ Polar Vectors vs. Axial Vectors (Rotational Cross Products) Distinguished! +4 Marks

✖ Linear displacement (r), velocity (v), momentum (p), force (F), and electric field (E) are polar (true) vectors, whereas ω, α, τ = r × F, and L = r × p are axial vectors!

Concept & Solution:
• Polar Vectors have a starting point or point of application along the direction of motion/push: Displacement (r), Velocity (v), Linear Momentum (p), Force (F), Electric Field (E).
• Axial Vectors (Pseudovectors) represent rotational effects and act along the axis of rotation via the cross product: ω, α, τ = r × F, L = r × p, and Magnetic Field (B).

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Cross Product Collinearity Condition (A × B = 0) JEE Main 2024 (Apr Shift 2)

⏱ JEE Standard Pace: 120s+4 / -1 Scale

Q8. If the two vectors A = (2 î − 3 ĵ + 4 k̂) and B = (−6 î + α ĵ + β k̂) are parallel (collinear) to each other, then what is the ordered pair of values (α, β)?

A (α = −9, β = −12) B (α = +9, β = −12) C (α = +9, β = +12) D (α = +6, β = −8)

✔ Proportional Components Ratio B = −3 A Mastered! +4 Marks

✖ For parallel/antiparallel vectors, components are proportional: −6/2 = α/(−3) = β/4 = −3 ⇒ α = +9 and β = −12!

Concept & Solution:
• Two vectors A and B are collinear (A × B = 0) if and only if their corresponding Cartesian components are in the same ratio:
• B_x / A_x = B_y / A_y = B_z / A_z.
• (−6) / 2 = α / (−3) = β / 4 = −3.
• Therefore, α = (−3)(−3) = +9 and β = 4(−3) = −12.

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Lami’s Theorem (Joined Strings with Two Weights at Two Knots) JEE Main 2025 (Jan Shift 2)

⏱ JEE Standard Pace: 120s+4 / -1 Scale

Q9. A knot P is tied to a ceiling string OP (making 30° with the vertical), a horizontal string PQ pulling to the right with tension T_PQ, and a vertical load of weight W = 30√3 N. What is the tension T_PQ in the horizontal string PQ required to maintain knot P in static equilibrium?

A 90 N B 60 N C 30 N D 15√3 N

✔ Tangent Relation T_PQ = W tan 30° = 30 N Mastered! +4 Marks

✖ By Lami’s Theorem at knot P, T_PQ / sin(150°) = W / sin(120°) ⇒ T_PQ = W tan 30° = 30√3 × (1/√3) = 30 N!

Concept & Solution:
• At knot P, three concurrent forces act in equilibrium: T_OP (at 30° to vertical), horizontal tension T_PQ, and downward weight W = 30√3 N.
• Vertical equilibrium: T_OP cos 30° = W.
• Horizontal equilibrium: T_PQ = T_OP sin 30°.
• Dividing the two gives: T_PQ = W tan 30° = 30√3 × (1 / √3) = 30 N.

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Dot Product Application (Instantaneous Power P = F · v) JEE Main 2023 (Apr Shift 2)

⏱ JEE Standard Pace: 120s+4 / -1 Scale

Q10. A particle moves under the action of a force F = (4 î + 3 ĵ − 2 k̂) N with an instantaneous velocity v = (2 î − 6 ĵ + c k̂) m s⁻¹. If the instantaneous power delivered by the force to the particle at that instant is zero, what is the value of c?

A +5 B +10 C −10 D −5

✔ Zero Power P = F · v = 0 Solved Cleanly! +4 Marks

✖ Instantaneous power is P = F · v = (4)(2) + (3)(−6) + (−2)(c) = 0 ⇒ 8 − 18 − 2c = 0 ⇒ −2c = 10 ⇒ c = −5!

Concept & Solution:
• Instantaneous power delivered by a force is the scalar product of force and velocity: P = F · v.
• Setting P = 0:
• (4)(2) + (3)(−6) + (−2)(c) = 0 ⇒ 8 − 18 − 2 c = 0.
• −10 − 2 c = 0 ⇒ c = −5.

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Vector Addition (Minimum Number of Non-Coplanar Vectors for Zero Sum) JEE Main 2022 (July Shift 1)

⏱ JEE Standard Pace: 120s+4 / -1 Scale

Q11. What is the minimum number of non-coplanar vectors (such that all of them do not lie in the same single plane) whose vector sum can be equal to the null (zero) vector?

A 3 B 4 C 2 D 5

✔ Coplanar (3) vs. Non-Coplanar (4) Zero-Sum Threshold Mastered! +4 Marks

✖ Three vectors that sum to zero (A + B + C = 0) must form a closed triangle and therefore MUST be coplanar! To have non-coplanar vectors summing to zero, you need at least 4 vectors!

Concept & Solution:
• Any two vectors A and B always define a plane, so their resultant (A + B) lies in that same plane.
• Consequently, a third vector C = −(A + B) that cancels them must also lie in the plane of A and B (so 3 vectors summing to zero are always coplanar).
• Therefore, at least 4 non-coplanar vectors are required to produce a zero resultant in 3D space.

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Cross Product Application (Lorentz Magnetic Force F = q(v × B)) JEE Main 2024 (Jan Shift 1)

⏱ JEE Standard Pace: 120s+4 / -1 Scale

Q12. A particle of charge q = 2 C moves with velocity v = (2 î + 3 ĵ) m s⁻¹ in a uniform magnetic field B = (4 ĵ + 3 k̂) T. What is the magnetic force vector F = q (v × B) acting on the particle?

A (18 î − 12 ĵ + 16 k̂) N B (9 î − 6 ĵ + 8 k̂) N C (18 î + 12 ĵ + 16 k̂) N D (−18 î + 12 ĵ − 16 k̂) N

✔ Cross Product v × B Multiplied by q = 2 C Mastered! +4 Marks

✖ Compute v × B = (2î + 3ĵ) × (4ĵ + 3k̂) = 8k̂ − 6ĵ + 9î = 9î − 6ĵ + 8k̂, then multiply by q = 2 C to get 18î − 12ĵ + 16k̂!

Concept & Solution:
• First compute v × B = (2 î + 3 ĵ) × (4 ĵ + 3 k̂):
• = 8 (î × ĵ) + 6 (î × k̂) + 12 (ĵ × ĵ) + 9 (ĵ × k̂)
• = 8 k̂ − 6 ĵ + 0 + 9 î = 9 î − 6 ĵ + 8 k̂.
• Multiplying by charge q = 2 C: F = 2 (9 î − 6 ĵ + 8 k̂) = (18 î − 12 ĵ + 16 k̂) N.

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Resolution of Vectors (Walking Displacements on Compass Grid) JEE Main 2023 (Jan Shift 1)

⏱ JEE Standard Pace: 120s+4 / -1 Scale

Q13. A person walks 10 m due North, then 20 m due East, and finally 30√2 m in the South-West direction (45° South of West). What is the net displacement vector of the person from the starting point (taking East as +î and North as +ĵ)?

A (−10 î + 20 ĵ) m (10 m West and 20 m North) B (+10 î − 20 ĵ) m (10 m East and 20 m South) C (−10 î − 20 ĵ) m (10 m West and 20 m South; Magnitude = 10√5 m) D (−20 î − 10 ĵ) m (20 m West and 10 m South)

✔ South-West Resolution (−30 î − 30 ĵ) Mastered! +4 Marks

✖ 30√2 m South-West resolves into −30 î − 30 ĵ m! Adding 10 ĵ + 20 î gives (20 − 30)î + (10 − 30)ĵ = −10 î − 20 ĵ m!

Concept & Solution:
• 1^st displacement: s₁ = +10 ĵ m.
• 2^nd displacement: s₂ = +20 î m.
• 3^rd displacement (30√2 m South-West): s₃ = −30√2 cos 45° î − 30√2 sin 45° ĵ = −30 î − 30 ĵ m.
• Net Displacement: s = s₁ + s₂ + s₃ = (20 − 30) î + (10 − 30) ĵ = (−10 î − 20 ĵ) m, with magnitude 10√5 m.

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Vector Triple Product (A × (B × C) BAC-CAB Identity) JEE Main 2025 (Apr Shift 1)

⏱ JEE Standard Pace: 120s+4 / -1 Scale

Q14. Given the vectors A = î + ĵ, B = ĵ + k̂, and C = k̂ + î, what is the vector triple product A × (B × C)?

A î + ĵ − 2 k̂ B −î + ĵ C î − ĵ D 0 (Null Vector)

✔ BAC-CAB Identity B(A·C) − C(A·B) Applied in Seconds! +4 Marks

✖ Use A × (B × C) = B(A · C) − C(A · B)! Since A · C = 1 and A · B = 1, A × (B × C) = B − C = (ĵ + k̂) − (k̂ + î) = −î + ĵ!

Concept & Solution:
• Using the vector triple product (BAC − CAB) identity:
• A × (B × C) = B (A · C) − C (A · B).
• Here, A · C = (î + ĵ) · (k̂ + î) = 1, and A · B = (î + ĵ) · (ĵ + k̂) = 1.
• Therefore, A × (B × C) = 1 B − 1 C = (ĵ + k̂) − (k̂ + î) = −î + ĵ.

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Set 33 Grand Finale (495-PYQ Milestone: Lami’s Minimum Tension by Varying Angle) JEE Main 2026 (Jan Shift 2)

⏱ JEE Standard Pace: 120s+4 / -1 Scale

Q15. A heavy lamp of weight W = 80 N is supported in equilibrium by two light cables OA and OB. Cable OA is fixed at an angle of 30° with the vertical (carrying tension T₁), while the inclination angle θ of cable OB with the horizontal can be adjusted to minimize its tension T₂. For what angle between cable OB and cable OA is tension T₂ minimum, and what are that minimum tension (T_2,min) and the corresponding tension T₁?

A 60° between OA & OB ; T_2,min = 40 N ; T₁ = 80 N B 90° between OA & OB ; T_2,min = 40√3 N ; T₁ = 40 N C 120° between OA & OB ; T_2,min = 40 N ; T₁ = 40√3 N D 90° between OA & OB ; T_2,min = 40 N ; T₁ = 40√3 N

✔ 495th Question Milestone Conquered! Lami’s Sine-Denominator Maximized at 90°! +4 Marks

✖ By Lami’s Theorem, T₂ / sin(150°) = W / sin(∠AOB) ⇒ T₂ = W sin 30° / sin(∠AOB) is MINIMUM when sin(∠AOB) = 1 (∠AOB = 90°)!

Concept & Solution:
• Angle between cable OA (30° to upward vertical) and downward weight W is 180° − 30° = 150°.
• Let the angle between cables OA and OB be φ. By Lami’s Theorem:
• T₂ / sin 150° = W / sin φ ⇒ T₂ = (W sin 30°) / sin φ = 40 / sin φ.
• Clearly, T₂ is minimum when sin φ = 1 ⇒ φ = 90° (cable OB is perpendicular to cable OA), giving T_2,min = 40 N and T₁ = W cos 30° = 40√3 N.

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Set 33 — 495-PYQ Series Completed!

Phenomenal milestone! You are just 1 set away from the 500-PYQ mark, with 495 unique, zero-dependency JEE Main PYQs across Sets 1–33.

Restart Review from Q1 →

IITJEE-MAINS Physics Previous Year Question Quiz Practice Set 33

TOPIC: Scalars vs Vectors, Addition, Resolution, Dot/Cross Products, and Lami’s Theorem

IIT-JEE MAINS PYQ Arena — Set 33

Set 33 — 495-PYQ Series Completed!

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