TOPIC: Scalars vs Vectors, Addition, Resolution, Dot/Cross Products, and Lami’s Theorem
IIT-JEE MAINS PYQ Arena — Set 34
Scalars vs Vectors, Addition, Resolution, Dot/Cross Products & Lami’s Theorem
Set 34 • 510-PYQ Milestone
Vector Addition (Ratio of Sum & Difference Magnitudes at 60°)
JEE Main 2024 (Jan Shift 1)
⏱ JEE Standard Pace: 120s
+4 / -1 Scale
Q1. Two vectors A and B have magnitudes in the ratio |A| : |B| = 2 : 1 and act at an angle of θ = 60° to each other. What is the ratio of the magnitude of their sum to the magnitude of their difference, |A + B| : |A − B|?
✔ Cosine Law Sum √(4 + 1 + 2) vs. Difference √(4 + 1 − 2) Mastered! +4 Marks
✖ Let |A| = 2k and |B| = k. With cos 60° = 1/2, |A + B| = k√(4 + 1 + 2) = k√7 and |A − B| = k√(4 + 1 − 2) = k√3!
Concept & Solution:• Let |A| = 2k and |B| = k, with cos 60° = 1 / 2.
• |A + B| = √(A2 + B2 + 2 A B cos 60°) = k √(4 + 1 + 2(2)(1)(1/2)) = k √7.
• |A − B| = √(A2 + B2 − 2 A B cos 60°) = k √(4 + 1 − 2(2)(1)(1/2)) = k √3.
• Therefore, |A + B| : |A − B| = √7 : √3.
Question 1 of 15
Lami’s Theorem (Horizontal Strut & Inclined Tie-Cable)
JEE Main 2025 (Jan Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q2. A load of weight W = 120 N is suspended from the tip B of a light horizontal strut AB hinged to a vertical wall at A. Tip B is supported by a light cable BC attached to the wall at C such that cable BC makes an angle of 30° with the horizontal strut AB, as shown below. Using Lami’s Theorem at B, what are the tension T in cable BC and the horizontal outward thrust FAB exerted by strut AB?
✔ Lami’s Equilibrium at Tip B (T sin 30° = W & FAB = T cos 30°) Mastered! +4 Marks
✖ At tip B, vertical balance gives T sin 30° = 120 N ⇒ T = 240 N, and horizontal balance gives FAB = T cos 30° = 120√3 N!
Concept & Solution:• At tip B, three concurrent forces are in equilibrium: cable tension T (up and left at 30° to horizontal), horizontal strut thrust FAB (to the right), and weight W = 120 N (downwards).
• Vertical equilibrium: T sin 30° = W = 120 N ⇒ T = 240 N.
• Horizontal equilibrium: FAB = T cos 30° = 240 × (√3 / 2) = 120√3 N.
Resolution of Vectors (Unit Vector Along Angle Bisector)
JEE Main 2024 (Apr Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q3. Given two vectors A = (2 î + 2 ĵ − k̂) and B = (6 î − 3 ĵ + 2 k̂), what is the unit vector that points along the internal angle bisector of vectors A and B?
✔ Angle Bisector Formula (Â + B̂) / |Â + B̂| Mastered! +4 Marks
✖ Since |A| = 3 ≠ |B| = 7, A + B does NOT bisect the angle! You must add their UNIT vectors: Â + B̂ = A/3 + B/7 = (7A + 3B)/21!
Concept & Solution:• A vector along the internal angle bisector of two unequal vectors A and B is directed along the sum of their unit vectors (Â + B̂).
• Here |A| = √(4 + 4 + 1) = 3, and |B| = √(36 + 9 + 4) = 7.
• Â + B̂ = A / 3 + B / 7 = (7 A + 3 B) / 21 = [ (14 + 18)î + (14 − 9)ĵ + (−7 + 6)k̂ ] / 21 = (32 î + 5 ĵ − k̂) / 21.
• Normalizing gives the unit bisector: (32 î + 5 ĵ − k̂) / √(322 + 52 + (−1)2) = (32 î + 5 ĵ − k̂) / √1050.
Cross Product Identity |(A + B) × (A − B)|
JEE Main 2023 (Jan Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q4. For any two vectors A and B, the cross product of their sum and difference, (A + B) × (A − B), is identically equal to which of the following expressions?
✔ Anti-Commutativity B × A = −(A × B) Respected! +4 Marks
✖ Expand carefully: A × A − A × B + B × A − B × B = 0 − A × B − A × B − 0 = −2(A × B) = 2(B × A)!
Concept & Solution:• Expanding using the distributive property of the cross product:
• (A + B) × (A − B) = (A × A) − (A × B) + (B × A) − (B × B).
• Since A × A = 0, B × B = 0, and B × A = −(A × B):
• (A + B) × (A − B) = −2 (A × B) = 2 (B × A).
500-PYQ Milestone Question! (Dot Product Angle Between Diagonals of a Cube)
JEE Main 2025 (Apr Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q5. [500th Question Milestone!] Consider a cube of side length a placed with one corner at the origin and its three adjacent edges along the positive x, y, and z axes. What is the acute angle θ between any two main body diagonals of the cube?
✔ Historic 500th Question Conquered! Cube Body Diagonals cos θ = 1/3! +4 Marks
✖ Take two body diagonals d1 = a(î + ĵ + k̂) and d2 = a(î + ĵ − k̂). Their dot product gives cos θ = (1 + 1 − 1) / (√3 × √3) = 1/3!
Concept & Solution:• Vector along the first main body diagonal (from (0,0,0) to (a,a,a)): d1 = a (î + ĵ + k̂), with |d1| = a√3.
• Vector along a second main body diagonal (from (0,0,a) to (a,a,0)): d2 = a (î + ĵ − k̂), with |d2| = a√3.
• Taking their dot product: cos θ = |d1 · d2| / (|d1| |d2|) = a2 (1 + 1 − 1) / (3 a2) = 1 / 3 ⇒ θ = cos−1(1 / 3).
Vector Addition (Condition for |A + B| < |A − B|)
JEE Main 2023 (Apr Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q6. For two non-zero vectors A and B inclined at an angle θ (where 0° ≤ θ ≤ 180°), the magnitude of their sum is strictly less than the magnitude of their difference (i.e., |A + B| < |A − B|) if and only if the angle θ lies in which range?
✔ Obtuse Angle Condition (cos θ < 0) Mastered! +4 Marks
✖ Squaring |A + B| < |A − B| gives 4|A||B| cos θ < 0 ⇒ cos θ < 0, which holds for obtuse angles (90° < θ ≤ 180°)!
Concept & Solution:• Squaring both sides of |A + B| < |A − B|:
• A2 + B2 + 2 A B cos θ < A2 + B2 − 2 A B cos θ.
• 4 A B cos θ < 0 ⇒ cos θ < 0.
• Therefore, the angle θ between A and B must be obtuse (90° < θ ≤ 180°).
Lami’s Theorem (Two Charged Pith Balls in Electrostatic Equilibrium)
JEE Main 2024 (Jan Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q7. Two small spheres having unequal masses m1 and m2 (with m1 > m2) and carrying charges q1 and q2 of the same sign are suspended from the same point by two insulating threads of equal length L. If the threads make angles θ1 and θ2 with the vertical in static equilibrium, how are θ1 and θ2 related?
✔ Newton’s 3rd Law Equal Electrostatic Repulsion + Lami’s Theorem Mastered! +4 Marks
✖ By Newton’s 3rd Law, the repulsive force Fe on both spheres is identical in magnitude! Heavier mass m1 > m2 deflects by a smaller angle (θ1 < θ2)!
Concept & Solution:• Regardless of the values of q1 and q2, the mutual electrostatic repulsion Fe has the exact same magnitude on both spheres.
• Since both threads have equal length L, the line joining the two spheres is tilted if m1 ≠ m2; taking moments about the suspension point (or Lami’s theorem for small angles):
• m1 g L sin θ1 = m2 g L sin θ2 ⇒ sin θ1 / sin θ2 = m2 / m1.
• Since m1 > m2, we have θ1 < θ2 (the heavier ball stays closer to the vertical).
Scalar Triple Product (Volume of a Parallelepiped)
JEE Main 2023 (Jan Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q8. Three coterminous edges of a parallelepiped are represented by the vectors A = (2 î − 3 ĵ + 4 k̂) m, B = (î + 2 ĵ − k̂) m, and C = (3 î − ĵ + 2 k̂) m. What is the volume of the parallelepiped?
✔ Volume V = |A · (B × C)| = |−7| = 7 m3 Mastered! +4 Marks
✖ Compute B × C = (4 − 1)î − (2 − (−3))ĵ + (−1 − 6)k̂ = 3î − 5ĵ − 7k̂, then |A · (B × C)| = |6 + 15 − 28| = 7 m3!
Concept & Solution:• The volume of a parallelepiped with coterminous edges A, B, C is the magnitude of their scalar triple product: V = |A · (B × C)|.
• First compute B × C: î(4 − 1) − ĵ(2 + 3) + k̂(−1 − 6) = 3 î − 5 ĵ − 7 k̂.
• Taking dot product with A: A · (B × C) = (2)(3) + (−3)(−5) + (4)(−7) = 6 + 15 − 28 = −7.
• Volume V = |−7| = 7 m3.
Resolution of Vectors (Inclined Pull Force on a Horizontal Block)
JEE Main 2022 (June Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q9. A block of mass m = 10 kg rests on a horizontal floor. A pulling force of magnitude F = 40 N is applied to the block at an angle of 30° above the horizontal, while in a second experiment the same 40 N force is applied as a push at 30° below the horizontal. Taking g = 10 m s−2, what are the normal reactions (Npull and Npush) exerted by the floor in the two cases?
✔ Pulling (mg − F sin θ = 80 N) vs. Pushing (mg + F sin θ = 120 N) Mastered! +4 Marks
✖ Vertical component F sin 30° = 40(0.5) = 20 N acts UPWARDS during a pull (reducing N to 100 − 20 = 80 N) and DOWNWARDS during a push (increasing N to 100 + 20 = 120 N)!
Concept & Solution:• Vertical component of force F: Fy = F sin 30° = 40 × (1 / 2) = 20 N.
• When pulling (Fy acts upwards): Npull = m g − F sin 30° = 100 − 20 = 80 N.
• When pushing (Fy acts downwards): Npush = m g + F sin 30° = 100 + 20 = 120 N.
Dot Product (Condition for Acute vs. Obtuse Angle Between Vectors)
JEE Main 2024 (Apr Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q10. Two vectors are given by A = (c î − 6 ĵ + 3 k̂) and B = (c î − 2 ĵ − 2 c k̂). For what set of real values of c do the vectors A and B make an obtuse angle with each other?
✔ Completing the Square A · B = c2 − 6c + 12 = (c − 3)2 + 3 > 0 Mastered! +4 Marks
✖ For an obtuse angle, we need A · B < 0! Here A · B = c2 + 12 − 6c = (c − 3)2 + 3, which is ALWAYS positive (≥ 3) for every real c!
Concept & Solution:• For an obtuse angle between A and B, their dot product must be negative: A · B < 0.
• Computing the dot product: A · B = (c)(c) + (−6)(−2) + (3)(−2 c) = c2 − 6 c + 12.
• Completing the square: A · B = (c − 3)2 + 3 ≥ +3 > 0 for all real values of c.
• Thus the angle between A and B is always acute, and there is no real value of c for which the angle is obtuse.
Vector Addition (Resultant of Three Concurrent Forces on Coordinate Grid)
JEE Main 2025 (Jan Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q11. Three coplanar forces act at the origin O in the x-y plane: F1 = 10 N along the positive x-axis (0°), F2 = 10√2 N at 45° in the first quadrant, and F3 = 20 N along the negative x-axis (180°). What are the magnitude and direction of the net resultant force R = F1 + F2 + F3?
✔ Component Resolution (10î + 10î + 10ĵ − 20î = +10ĵ) Mastered! +4 Marks
✖ Resolve F2 = 10√2(cos 45° î + sin 45° ĵ) = 10î + 10ĵ N! Adding F1 = 10î and F3 = −20î cancels the x-component completely, leaving +10ĵ N!
Concept & Solution:• F1 = +10 î N.
• F2 = 10√2 cos 45° î + 10√2 sin 45° ĵ = (10 î + 10 ĵ) N.
• F3 = −20 î N.
• Adding all three vectors: R = (10 + 10 − 20) î + 10 ĵ = +10 ĵ N (10 N along the positive y-axis).
Scalars vs. Vectors (Stress, Pressure & Area Vector)
JEE Main 2023 (Apr Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q12. In physics, how are Hydrostatic Pressure (P) in a fluid, Infinitesimal Surface Area Element (dA), and Mechanical Stress (σij) inside an anisotropic solid classified respectively?
✔ Scalar (Rank 0) vs. Vector (Rank 1) vs. Tensor (Rank 2) Distinction Mastered! +4 Marks
✖ Hydrostatic pressure acts equally in all directions (Scalar), dA = n̂ dA is a Vector directed along the outward normal, and Stress depends on both force direction and surface normal (Rank-2 Tensor)!
Concept & Solution:• Hydrostatic Pressure (P) at a point in a fluid has no unique direction (Pascal’s Law) and is a Scalar (where dF = P dA).
• Elementary Area (dA = n̂ dA) is a Vector directed perpendicular (outward normal) to the surface.
• Mechanical Stress (σij) requires two directions (direction of force and normal to the plane) and is a Second-Rank Tensor.
Cross Product & Dot Product Combined System
JEE Main 2024 (Jan Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q13. Given a fixed non-zero vector A, if another vector B satisfies both A · B = 0 and A × B = 0 simultaneously, then what must be true about vector B?
✔ Lagrange’s Identity |A|2|B|2 = (A·B)2 + |A×B|2 = 0 Mastered! +4 Marks
✖ Since sin θ and cos θ can never be zero simultaneously for any angle θ, and |A| ≠ 0, |B| must be zero!
Concept & Solution:• Using Lagrange’s Identity: |A|2 |B|2 = (A · B)2 + |A × B|2.
• Since A · B = 0 and A × B = 0, the right-hand side is 0 + 0 = 0.
• Since A is a non-zero vector (|A| ≠ 0), we must have |B| = 0 ⇒ B = 0 (Null Vector).
Lami’s Theorem (Ring on Smooth Vertical Circular Hoop)
JEE Main 2025 (Apr Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q14. A small smooth ring of weight W = 50 N is threaded onto a fixed frictionless vertical circular hoop of radius R. The ring is held in static equilibrium at an angular position of 60° below the highest point of the hoop by a light string tied to the highest point A of the hoop. What are the tension T in the string and the normal reaction N exerted by the circular hoop on the ring?
✔ Equilateral Chord Triangle + Lami’s 120° Symmetry Mastered! +4 Marks
✖ Since central angle ∠AOP = 60° and OA = OP = R, triangle OAP is equilateral! Thus both N (along OP) and T (along PA) make 60° with the upward vertical (and 120° with each other), giving T = N = W = 50 N!
Concept & Solution:• Let O be the center of the hoop, A be the highest point (vertical radius OA), and P be the ring’s position with ∠AOP = 60°.
• Since OA = OP = R and ∠AOP = 60°, triangle OAP is equilateral.
• Thus the radial normal force N (along OP) makes 60° with the upward vertical, and the string tension T (along PA) makes 60° with the upward vertical on the opposite side (so ∠ between T and N is 120°).
• By Lami’s 120° symmetry: T = N = W = 50 N.
Set 34 Grand Finale (510-PYQ Milestone: Solving Vector Equation A · X = c & A × X = B)
JEE Main 2026 (Jan Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q15. Let A be a known non-zero vector, B be a known vector perpendicular to A, and k be a known scalar. If an unknown vector X satisfies both A · X = k and A × X = B simultaneously, then what is the unique expression for vector X in terms of A, B, and k?
✔ 510th Question Milestone Conquered! Cross with A & BAC-CAB Identity Mastered! +4 Marks
✖ Take the cross product of A × X = B with A from the left: A × (A × X) = A × B ⇒ A(A · X) − X(A · A) = A × B ⇒ kA − |A|2X = A × B!
Concept & Solution:• Taking the cross product of A × X = B with A on the left:
• A × (A × X) = A × B.
• Expanding the left-hand side via the BAC-CAB identity: A (A · X) − X (A · A) = A × B.
• Substituting A · X = k and A · A = |A|2:
• k A − |A|2 X = A × B ⇒ X = [ k A − (A × B) ] / |A|2 = [ k A + (B × A) ] / |A|2.
510-PYQ Mega-Milestone Conquered!
Historic milestone! With Set 34 complete, you have crossed 500 questions and now have 510 unique, zero-dependency JEE Main PYQs across Sets 1–34.
🏆 Set 34 Live Vectors Tracker (510-PYQ Milestone)