TOPIC: Scalars vs Vectors, Addition, Resolution, Dot/Cross Products, and Lami’s Theorem
IIT-JEE MAINS PYQ Arena — Set 35
Scalars vs Vectors, Addition, Resolution, Dot/Cross Products & Lami’s Theorem
Set 35 • 525-PYQ Series
Vector Addition (Resultant Magnitude When Angle is Halved)
JEE Main 2024 (Jan Shift 2)
⏱ JEE Standard Pace: 120s
+4 / -1 Scale
Q1. Two forces of equal magnitude F act at a point at an angle of 120° with each other and produce a resultant of magnitude R1. If the angle between the two forces is halved (to 60°) without changing their magnitudes, the new resultant has magnitude R2. What is the ratio (R2 : R1)?
✔ Equal-Vector Shortcut R = 2 F cos(θ/2) Applied Instantly! +4 Marks
✖ At θ = 120°, R1 = 2F cos(60°) = F; at θ = 60°, R2 = 2F cos(30°) = √3 F, so R2 : R1 = √3 : 1!
Concept & Solution:• For two equal forces F at angle θ, the resultant magnitude is R = 2 F cos(θ / 2).
• When θ1 = 120°: R1 = 2 F cos(60°) = F.
• When θ2 = 60°: R2 = 2 F cos(30°) = √3 F.
• Ratio R2 : R1 = √3 : 1.
Question 1 of 15
Lami’s Theorem (Two Identical Smooth Spheres in a Cylinder)
JEE Main 2025 (Jan Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q2. A smooth sphere of weight W = 80 N and radius R is held in static equilibrium on a frictionless inclined plane of inclination θ = 30° by a vertical smooth stop wall fixed perpendicular to the horizontal ground, as shown below. Using Lami’s Theorem, what are the normal reaction N1 exerted by the inclined plane and the horizontal normal reaction N2 exerted by the vertical stop wall?
✔ Vertical vs. Perpendicular Stop Wall Distinction (N1 cos 30° = W) Mastered! +4 Marks
✖ Since the stop wall is VERTICAL, its normal force N2 is purely HORIZONTAL! Thus N1 cos 30° = W = 80 N ⇒ N1 = 160/√3 N and N2 = N1 sin 30° = 80/√3 N!
Concept & Solution:• Since the stop wall is vertical, its normal reaction N2 is purely horizontal.
• The normal reaction N1 from the 30° incline makes an angle of 30° with the upward vertical.
• By Lami’s Theorem (or vertical equilibrium): N1 cos 30° = W = 80 N ⇒ N1 = 160 / √3 N.
• Horizontal equilibrium: N2 = N1 sin 30° = W tan 30° = 80 / √3 N.
Dot Product (Finding Unknown Vector in 2D Plane)
JEE Main 2024 (Apr Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q3. A vector B lies in the x-y plane, has a magnitude of |B| = 10, and is perpendicular to the vector A = (3 î − 4 ĵ). If the x-component of vector B is positive, what is the vector B?
✔ 2D Orthogonal Vector Swap (x, y) → (−y, x) Mastered! +4 Marks
✖ In 2D, any vector perpendicular to (3î − 4ĵ) is proportional to (4î + 3ĵ)! Since |(4î + 3ĵ)| = 5, scaling to magnitude 10 gives 2(4î + 3ĵ) = 8î + 6ĵ!
Concept & Solution:• In the x-y plane, a vector perpendicular to A = ax î + ay ĵ is parallel to ±(−ay î + ax ĵ).
• For A = 3 î − 4 ĵ, the perpendicular direction with positive x-component is along 4 î + 3 ĵ (check dot product: 3(4) − 4(3) = 0 ✔).
• Unit vector: n̂ = (4 î + 3 ĵ) / 5.
• Multiplying by magnitude |B| = 10 gives: B = 10 × (4 î + 3 ĵ) / 5 = 8 î + 6 ĵ.
Cross Product Application (Perpendicular Distance of Point from Line of Action)
JEE Main 2025 (Apr Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q4. A force F = (3 î + 4 ĵ) N acts at a point P whose position vector relative to the origin O is r = (2 î + ĵ) m. What are the torque vector (τ = r × F) about the origin O and the perpendicular moment arm (lever arm d⊥) of the line of action of force F from the origin O?
✔ Moment Arm Formula d⊥ = |r × F| / |F| = 5 / 5 = 1.0 m Mastered! +4 Marks
✖ τ = r × F = (2(4) − 1(3))k̂ = +5k̂ N·m! Since |τ| = |F| d⊥ and |F| = 5 N, the moment arm is d⊥ = 5 / 5 = 1.0 m!
Concept & Solution:• Torque about origin O: τ = r × F = (2 î + ĵ) × (3 î + 4 ĵ) = (8 − 3) k̂ = +5 k̂ N·m.
• Magnitude of force: |F| = √(32 + 42) = 5 N.
• Since |τ| = r⊥ |F| = d⊥ |F|, the perpendicular lever arm from O to the line of action of F is:
• d⊥ = |r × F| / |F| = 5 N·m / 5 N = 1.0 m.
Vector Addition (Centroid & Median Vector Sum of Triangle)
JEE Main 2023 (Jan Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q5. In any triangle ABC, let D, E, and F be the midpoints of sides BC, CA, and AB respectively, and let G be the centroid of the triangle. What are the values of the vector sums (AD + BE + CF) and (GA + GB + GC)?
✔ Sum of Medians of a Triangle = 0 Mastered! +4 Marks
✖ Since AD = (AB + AC)/2, BE = (BC + BA)/2, and CF = (CA + CB)/2, adding all three cancels every pair of opposite side vectors to give 0!
Concept & Solution:• Median vectors: AD = (AB + AC) / 2, BE = (BC + BA) / 2, and CF = (CA + CB) / 2.
• Adding all three medians: AD + BE + CF = [ (AB + BA) + (BC + CB) + (CA + AC) ] / 2 = 0 (Null Vector).
• Since centroid G divides each median in a 2 : 1 ratio (GA = −2/3 AD, etc.), we also have GA + GB + GC = 0.
Resolution of Vectors (Force Components Along & Normal to Incline)
JEE Main 2024 (Jan Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q6. A horizontal force of magnitude F = 40 N pushes a block of weight W = 40√3 N resting on a smooth inclined plane of inclination θ = 30° with the horizontal. What is the net force acting on the block parallel to the inclined plane?
✔ Horizontal Force Resolution F cos θ vs. W sin θ Mastered! +4 Marks
✖ Component of horizontal force F up the incline is F cos 30° = 40(√3/2) = 20√3 N, and component of weight W down the incline is W sin 30° = 40√3(1/2) = 20√3 N!
Concept & Solution:• Component of horizontal force F directed up the incline: F∥,up = F cos 30° = 40 × (√3 / 2) = 20√3 N.
• Component of weight W directed down the incline: W∥,down = W sin 30° = 40√3 × (1 / 2) = 20√3 N.
• Net force along the incline = 20√3 − 20√3 = 0 N (exact static equilibrium!).
Dot Product (Angle Between Face Diagonals of a Cube)
JEE Main 2023 (Apr Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q7. Consider a cube of edge length a with one vertex at the origin O(0, 0, 0) and its three coterminous edges along the positive x, y, and z axes. What is the angle between two face diagonals drawn from the origin O on two adjacent faces (i.e., along d1 = a(î + ĵ) and d2 = a(ĵ + k̂))?
✔ Equilateral Triangle of Three Face Diagonals (θ = 60°) Mastered! +4 Marks
✖ Dot product d1 · d2 = a2(0 + 1 + 0) = a2, and |d1| = |d2| = a√2, so cos θ = a2 / (2a2) = 1/2 ⇒ θ = 60°!
Concept & Solution:• Using the dot product: cos θ = (d1 · d2) / (|d1| |d2|).
• Here d1 · d2 = a2 (î + ĵ) · (ĵ + k̂) = a2, and |d1| = |d2| = a√2.
• Therefore, cos θ = a2 / (2 a2) = 1 / 2 ⇒ θ = 60° (geometrically, the three face diagonals form an equilateral triangle of side a√2!).
Vector Addition (Condition for |A + B|2 = |A|2 + |B|2 − |A||B|)
JEE Main 2022 (July Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q8. The resultant R = A + B of two non-zero vectors A and B satisfies the relation R2 = A2 + B2 − A B. What is the angle θ between vectors A and B?
✔ Cosine Law Comparison 2 A B cos θ = −A B ⇒ cos θ = −1/2 Mastered! +4 Marks
✖ Compare R2 = A2 + B2 + 2AB cos θ with A2 + B2 − AB: 2 cos θ = −1 ⇒ cos θ = −1/2 ⇒ θ = 120°!
Concept & Solution:• By the Parallelogram Law of vector addition: R2 = A2 + B2 + 2 A B cos θ.
• Comparing with the given relation R2 = A2 + B2 − A B:
• 2 A B cos θ = −A B ⇒ cos θ = −1 / 2 ⇒ θ = 120° (2π / 3 rad).
Cross Product Identity (A × B = B × C = C × A)
JEE Main 2025 (Apr Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q9. If three non-zero vectors A, B, and C satisfy the cyclic cross-product equality A × B = B × C = C × A ≠ 0, then what must the vector sum (A + B + C) be equal to?
✔ Cross-Product Factoring A × (A + B + C) = 0 & B × (A + B + C) = 0 Mastered! +4 Marks
✖ From A × B = C × A = −(A × C), we get A × (B + C) = 0 ⇒ A × (A + B + C) = 0 and B × (A + B + C) = 0! Since A and B are non-collinear, A + B + C = 0!
Concept & Solution:• From A × B = C × A = −(A × C), we have A × B + A × C = 0.
• Since A × A = 0, adding A × A gives: A × (A + B + C) = 0.
• Similarly, from B × C = A × B = −(B × A), we obtain: B × (A + B + C) = 0.
• Since A × B ≠ 0 (A and B are non-collinear), no non-zero vector can be parallel to both A and B simultaneously; hence A + B + C = 0.
Lami’s Theorem (Maximum Weight Supported Before String Breaks)
JEE Main 2024 (Apr Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q10. A weight W is suspended from a knot O tied to two ceiling strings OA and OB making angles of 60° and 30° with the horizontal ceiling respectively. Both strings OA and OB are made of the same material and have an identical breaking tension of Tmax = 100 N. If the suspended weight W is gradually increased, which string will break first, and what is the maximum weight (Wmax) that can be safely suspended?
✔ Steeper String Carries Greater Tension (TOA = W cos 30° > TOB = W sin 30°) Mastered! +4 Marks
✖ Since TOA = W sin 60° = (√3/2)W and TOB = W sin 30° = 0.5W, the steeper string OA always carries more tension and hits 100 N first when Wmax = 200/√3 N!
Concept & Solution:• Since 60° + 30° = 90°, the two strings are perpendicular to each other.
• By Lami’s Theorem: TOA = W sin 60° = (√3 / 2) W and TOB = W sin 30° = (1 / 2) W.
• Since TOA > TOB, the steeper string OA (at 60° with horizontal) reaches Tmax = 100 N first!
• Setting TOA = (√3 / 2) Wmax = 100 N gives Wmax = 200 / √3 N.
Resolution of Vectors (Reflected Velocity Vector Off Inclined Wall)
JEE Main 2025 (Jan Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q11. A smooth ball moving with initial velocity v1 = 10 î m s−1 strikes a fixed smooth wall whose outward normal unit vector is n̂ = −(î − ĵ) / √2 (i.e., a wall inclined at 45° to the x-axis). If the collision is perfectly elastic (so the velocity component along n̂ reverses sign while the tangential component remains unchanged), what is the final velocity vector v2 of the ball?
✔ Vector Reflection Formula v2 = v1 − 2(v1 · n̂)n̂ Mastered! +4 Marks
✖ Use the universal reflection vector formula v2 = v1 − 2(v1 · n̂)n̂ = 10î − 2(−10/√2)[−(î − ĵ)/√2] = 10î − 10(î − ĵ) = +10ĵ m s−1!
Concept & Solution:• Universal elastic reflection formula for any normal unit vector n̂:
• v2 = v1 − 2 (v1 · n̂) n̂.
• Here v1 · n̂ = 10 î · [ −(î − ĵ) / √2 ] = −10 / √2.
• Therefore, v2 = 10 î − 2 (−10 / √2) [ −(î − ĵ) / √2 ] = 10 î − 10 (î − ĵ) = +10 ĵ m s−1.
Dot Product & Cosine Law in Vector Triangle
JEE Main 2023 (Jan Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q12. Three vectors A, B, and C satisfy the relation A + B + C = 0. If their magnitudes are |A| = 3, |B| = 5, and |C| = 7, what is the angle θ between vectors A and B?
✔ Isolate Third Vector |A + B|2 = |−C|2 = 49 Mastered! +4 Marks
✖ Isolate C first: A + B = −C ⇒ |A|2 + |B|2 + 2|A||B| cos θ = |C|2 ⇒ 9 + 25 + 30 cos θ = 49 ⇒ cos θ = 1/2 ⇒ θ = 60°!
Concept & Solution:• Since we want the angle between A and B, isolate C on the other side: A + B = −C.
• Squaring both sides (taking dot product with itself):
• |A|2 + |B|2 + 2 |A| |B| cos θ = |C|2.
• 32 + 52 + 2(3)(5) cos θ = 72 ⇒ 34 + 30 cos θ = 49 ⇒ 30 cos θ = 15 ⇒ cos θ = 1 / 2 ⇒ θ = 60°.
Cross Product (Condition for Vector to Lie in Plane of Two Vectors)
JEE Main 2024 (Jan Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q13. What is the unit vector that lies in the plane of vectors A = (î + 2 ĵ − k̂) and B = (î + ĵ − 2 k̂) and is simultaneously perpendicular to vector C = (2 î − ĵ + k̂)?
✔ Coplanar & Orthogonal Vector C × (A × B) = ±(ĵ − k̂) Mastered! +4 Marks
✖ Any vector in the plane of A and B is of the form A − B = ĵ + k̂? Wait: check dot product with C (2î − ĵ + k̂): (ĵ − k̂) · (2î − ĵ + k̂) = −1 − 1 ≠ 0? No: A − B = (î + 2ĵ − k̂) − (î + ĵ − 2k̂) = ĵ + k̂, and (ĵ + k̂) · (2î − ĵ + k̂) = −1 + 1 = 0!
Concept & Solution:• Any vector in the plane of A and B is a linear combination P = λ A + μ B (or parallel to C × (A × B)).
• Notice that subtracting B from A gives A − B = (1 − 1)î + (2 − 1)ĵ + (−1 − (−2))k̂ = ĵ + k̂, whose dot product with C = (2 î − ĵ + k̂) is 0 − 1 + 1 = 0!
• (Using C · (A + μ B) = (2 − 2 − 1) + μ(2 − 1 − 2) = −1 − μ = 0 ⇒ μ = −1 ⇒ P = A − B).
• Normalizing gives magnitude √2.
Vector Addition (Resultant When One Vector is Reversed)
JEE Main 2025 (Apr Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q14. Two forces P and Q act at a point and have a resultant R1. If force Q is reversed in direction, the new resultant R2 = P − Q is observed to be perpendicular to the original resultant R1 = P + Q. What does this imply about the magnitudes of P and Q, and what is the value of (R12 + R22)?
✔ Orthogonal Diagonals (P + Q)·(P − Q) = P2 − Q2 = 0 & Parallelogram Law Mastered! +4 Marks
✖ Since R1 ⊥ R2, (P + Q) · (P − Q) = P2 − Q2 = 0 ⇒ |P| = |Q|, and R12 + R22 = 2(P2 + Q2) = 4P2!
Concept & Solution:• Since R1 = P + Q is perpendicular to R2 = P − Q:
• R1 · R2 = (P + Q) · (P − Q) = |P|2 − |Q|2 = 0 ⇒ |P| = |Q|.
• By the Parallelogram Identity:
• R12 + R22 = |P + Q|2 + |P − Q|2 = 2 (|P|2 + |Q|2) = 4 |P|2.
Set 35 Grand Finale (525-PYQ Milestone: Lami’s Equilibrium of Bead on Smooth Wire Circle)
JEE Main 2026 (Apr Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q15. Three concurrent coplanar forces F1, F2, and F3 keep a particle in static equilibrium. Force F1 = 60 N acts due East (+î), Force F2 acts at 60° North of West, and Force F3 acts at 30° South of West. Using Lami’s Theorem, what are the magnitudes of F2 and F3?
✔ 525th Question Milestone Conquered! Lami’s Angles (90°, 120°, 150°) Solved! +4 Marks
✖ Angle between F2 (60° N of W) and F3 (30° S of W) is 90°! Angle opposite F2 is 180° − 30° = 150°, and opposite F3 is 180° − 60° = 120°!
Concept & Solution:• Angle between F2 (60° N of W) and F3 (30° S of W) is 60° + 30° = 90°.
• Angle between F1 (East) and F3 (30° S of W) is 180° − 30° = 150° (opposite to F2).
• Angle between F1 (East) and F2 (60° N of W) is 180° − 60° = 120° (opposite to F3).
• By Lami’s Theorem: F2 / sin 150° = F3 / sin 120° = F1 / sin 90° = 60 N.
• Therefore, F2 = 60 × (1 / 2) = 30 N and F3 = 60 × (√3 / 2) = 30√3 N.
Set 35 — 525-PYQ Series Completed!
Phenomenal achievement! You now have 525 unique, zero-dependency JEE Main PYQs across Sets 1–35. Check your Set 35 score below or click any question pill above to review.
🧭 Set 35 Live Vectors Tracker (525-PYQ Series)