TOPIC: Scalars vs Vectors, Addition, Resolution, Dot/Cross Products, and Lami’s Theorem
IIT-JEE MAINS PYQ Arena — Set 30
Scalars vs Vectors, Addition, Resolution, Dot/Cross Products & Lami’s Theorem
Set 30 • 450-PYQ Milestone
Vector Addition (Maximum & Minimum Resultant Ratio)
JEE Main 2024 (Jan Shift 2)
⏱ JEE Standard Pace: 120s
+4 / -1 Scale
Q1. The ratio of the maximum possible resultant to the minimum possible resultant of two vectors P and Q (where |P| > |Q|) is 3 : 1. What is the ratio of their individual magnitudes (|P| : |Q|), and what will be the magnitude of their resultant when they act at 90° to each other?
✔ Componendo-Dividendo on (P + Q)/(P − Q) = 3/1 Mastered! +4 Marks
✖ Rmax / Rmin = (P + Q) / (P − Q) = 3 / 1 ⇒ P / Q = (3 + 1) / (3 − 1) = 2 : 1!
Concept & Solution:• Maximum resultant: Rmax = P + Q; Minimum resultant: Rmin = P − Q.
• Given (P + Q) / (P − Q) = 3 / 1 ⇒ P + Q = 3 P − 3 Q ⇒ 2 P = 4 Q ⇒ P : Q = 2 : 1.
• When acting at θ = 90°: R90° = √(P2 + Q2) = √((2Q)2 + Q2) = √5 |Q|.
Question 1 of 15
Lami’s Theorem (Sphere Resting Against Smooth Wall)
JEE Main 2025 (Jan Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q2. A uniform smooth sphere of weight W = 60 N and radius R rests against a frictionless vertical wall, held in equilibrium by a light string of length L = R attached to a point on its surface and fixed to a point on the wall, as shown below. What are the tension T in the string and the normal reaction N exerted by the vertical wall on the sphere?
✔ sin θ = R / (R + L) = 1/2 ⇒ θ = 30° + Lami’s Theorem Mastered! +4 Marks
✖ The line of the string passes through the center of the sphere with hypotenuse R + L = 2R and horizontal leg R, so sin θ = R/(2R) = 1/2 (θ = 30° with vertical wall)!
Concept & Solution:• Since the wall and sphere are smooth, the string’s line of action passes through the center C of the sphere.
• In the right triangle formed with the wall: hypotenuse = R + L = 2R, horizontal leg = R ⇒ sin θ = R / (2R) = 1 / 2 ⇒ θ = 30° (angle of string with the vertical wall).
• By Lami’s Theorem (or equilibrium): T cos 30° = W = 60 N ⇒ T = 120 / √3 = 40√3 N.
• Normal reaction from wall: N = T sin 30° = 40√3 × (1 / 2) = 20√3 N.
Resolution of Vectors (Unit Vector Reflected from Surface)
JEE Main 2024 (Apr Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q3. A ray of light travelling along the unit vector ê1 = (1 / 2) î − (√3 / 2) ĵ strikes a plane mirror lying along the x-axis (in the x-z plane) and gets reflected. What is the unit vector ê2 along the reflected ray, and what is the magnitude of the change in the unit vector |ê2 − ê1|?
✔ Normal Component Reversal (−ĵ → +ĵ) Mastered! +4 Marks
✖ Reflection from the x-axis leaves the parallel component (+1/2 î) unchanged and reverses the normal component (−√3/2 ĵ → +√3/2 ĵ)!
Concept & Solution:• Upon reflection from the x-axis (normal along ĵ), the component parallel to the mirror (î) remains unchanged while the perpendicular component (ĵ) reverses sign:
• ê2 = (1 / 2) î + (√3 / 2) ĵ.
• Change in unit vector: ê2 − ê1 = 2 × (√3 / 2) ĵ = √3 ĵ ⇒ |ê2 − ê1| = √3.
Dot & Cross Products (Lagrange’s Identity)
JEE Main 2023 (Jan Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q4. For any two vectors A and B with magnitudes |A| = 3 and |B| = 4, what is the value of the expression |A × B|2 + (A · B)2?
✔ Lagrange’s Vector Identity |A × B|2 + (A · B)2 = A2 B2 Mastered! +4 Marks
✖ Since sin2θ + cos2θ = 1, (AB sin θ)2 + (AB cos θ)2 = A2 B2 = 32 × 42 = 9 × 16 = 144!
Concept & Solution:• Using |A × B| = A B sin θ and A · B = A B cos θ:
• |A × B|2 + (A · B)2 = A2 B2 (sin2θ + cos2θ) = A2 B2.
• Substituting A = 3 and B = 4 gives: 32 × 42 = 9 × 16 = 144.
Vector Addition (Polygon Law & Regular Hexagon Sum)
JEE Main 2024 (Jan Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q5. If ABCDEF is a regular hexagon with O as its center, what is the resultant of the five vectors drawn from vertex A to the remaining five vertices: AB + AC + AD + AE + AF, expressed in terms of the diameter vector AD (or radius vector AO)?
✔ Opposite-Side Symmetry Pairing (AB + AE = AD, AC + AF = AD) Mastered! +4 Marks
✖ In a regular hexagon ABCDEF, CD is parallel and equal to AF, so AC + AF = AC + CD = AD! Similarly AB + AE = AD, giving AD + AD + AD = 3 AD = 6 AO!
Concept & Solution:• In a regular hexagon ABCDEF, opposite sides are equal and parallel: AE = BD and AF = CD.
• Pairing symmetrically:
• AB + AE = AB + BD = AD.
• AC + AF = AC + CD = AD.
• Therefore, AB + AC + AD + AE + AF = AD + AD + AD = 3 AD = 6 AO.
Resolution of Vectors (Component Perpendicular to Another Vector)
JEE Main 2025 (Apr Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q6. Given two vectors A = (4 î + 6 ĵ) and B = (î + ĵ), what is the vector component of A that is perpendicular to vector B and lies in the plane of A and B?
✔ Orthogonal Decomposition A⊥ = A − A∥ Mastered! +4 Marks
✖ First find the parallel component A∥ = [(A · B) / |B|2] B = (10 / 2)(î + ĵ) = 5î + 5ĵ, then subtract it from A (4î + 6ĵ)!
Concept & Solution:• Any vector A can be decomposed into components parallel and perpendicular to B: A = A∥ + A⊥.
• Parallel vector component: A∥ = [ (A · B) / |B|2 ] B = [ (4 + 6) / (12 + 12) ] (î + ĵ) = 5 î + 5 ĵ.
• Perpendicular vector component: A⊥ = A − A∥ = (4 î + 6 ĵ) − (5 î + 5 ĵ) = −î + ĵ (check: (−î + ĵ) · (î + ĵ) = −1 + 1 = 0 ✔).
Cross Product & Linear Velocity v = ω × r
JEE Main 2024 (Apr Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q7. A rigid body rotates with an angular velocity vector ω = (2 î − 3 ĵ + k̂) rad s−1 about an axis passing through the origin. What is the instantaneous linear velocity vector (v = ω × r) of a point on the body located at position vector r = (î + 2 ĵ − 2 k̂) m?
✔ Order of Cross Product v = ω × r (NOT r × ω) Respected! +4 Marks
✖ Remember: Linear velocity is v = ω × r (whereas angular momentum/torque have r first)! Reversing the order flips all signs!
Concept & Solution:• Using v = ω × r:
• vx = (−3)(−2) − (1)(2) = 6 − 2 = +4.
• vy = −[ (2)(−2) − (1)(1) ] = −(−4 − 1) = +5.
• vz = (2)(2) − (−3)(1) = 4 + 3 = +7.
• Therefore, v = (4 î + 5 ĵ + 7 k̂) m s−1.
Lami’s Theorem (Symmetric Clothesline Sag Tension)
JEE Main 2022 (July Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q8. A weight W is suspended from the midpoint of a light rope stretched between two hooks at the same horizontal level. Each half of the rope makes an angle θ with the horizontal. What is the tension T in the rope, and what happens to the tension T as the rope is pulled to become completely horizontal (θ → 0°)?
✔ Vertical Equilibrium 2 T sin θ = W ⇒ T = W / (2 sin θ) Mastered! +4 Marks
✖ Since θ is measured with the HORIZONTAL, the upward vertical components are 2 T sin θ = W ⇒ T = W / (2 sin θ) → ∞ as θ → 0°!
Concept & Solution:• Resolving the two symmetric tensions vertically (or applying Lami’s Theorem with angle 180° − 2θ between the two rope segments):
• 2 T sin θ = W ⇒ T = W / (2 sin θ).
• For the rope to become completely horizontal (θ → 0°, so sin θ → 0), the required tension T → ∞ (hence any real rope with a weight at its center must sag).
Vector Addition (N Equal Vectors Separated by Angle 2π/N)
JEE Main 2023 (Jan Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q9. Twelve coplanar forces, each of equal magnitude F = 10 N, act on a body at a point such that the angle between any two adjacent forces is 30° (forming a symmetric 12-spoke star). If one of the 12 forces is suddenly removed, what is the magnitude and direction of the net resultant of the remaining 11 forces?
✔ Closed Regular Polygon Zero-Sum Principle Mastered! +4 Marks
✖ All 12 symmetric forces sum to ZERO (F1 + F2 + … + F12 = 0), so the sum of the remaining 11 forces equals −F12!
Concept & Solution:• Since N = 12 forces of equal magnitude F are symmetrically spaced by 360° / 12 = 30°, they form a closed regular 12-sided polygon:
• F1 + F2 + … + F11 + F12 = 0.
• When F12 is removed, the resultant of the remaining 11 forces is:
• R11 = −F12, which has magnitude 10 N directed opposite to the removed force.
Dot Product Application (Work Done W = F · Δr & Power P = F · v)
JEE Main 2024 (Apr Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q10. A constant force F = (3 î + 4 ĵ − 5 k̂) N acts on a particle and displaces it from initial position r1 = (î + 2 ĵ + k̂) m to final position r2 = (3 î + 4 ĵ − k̂) m. What is the work done (W = F · Δr) by the force, and what is the cosine of the angle (cos θ) between F and the displacement vector Δr?
✔ Displacement Δr = r2 − r1 & Dot Product W = 24 J Mastered! +4 Marks
✖ Δr = r2 − r1 = 2î + 2ĵ − 2k̂ m! Then W = F · Δr = 3(2) + 4(2) + (−5)(−2) = 6 + 8 + 10 = 24 J!
Concept & Solution:• Displacement vector: Δr = r2 − r1 = (3 − 1)î + (4 − 2)ĵ + (−1 − 1)k̂ = (2 î + 2 ĵ − 2 k̂) m.
• Work Done W = F · Δr = (3)(2) + (4)(2) + (−5)(−2) = 6 + 8 + 10 = 24 J.
• Magnitudes: |F| = √(9 + 16 + 25) = √50 = 5√2 N; |Δr| = √(4 + 4 + 4) = 2√3 m.
• cos θ = W / (|F| |Δr|) = 24 / (10√6) = 12 / (5√6) = 2√6 / 5.
Non-Orthogonal Resolution of a Vector
JEE Main 2025 (Jan Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q11. A force of magnitude F = 20 N is resolved into two coplanar components F1 and F2 making angles of α = 30° and β = 45° on opposite sides of the force F. Using the sine rule of vector resolution, what is the magnitude of component F1 (which makes 30° with F)?
✔ Non-Orthogonal Resolution Formula F1 = F sin β / sin(α + β) Mastered! +4 Marks
✖ Since α + β = 30° + 45° = 75° ≠ 90°, you CANNOT use F cos 30°! Use the sine rule: F1 / sin 45° = F / sin 75°!
Concept & Solution:• Since α + β = 30° + 45° = 75° (non-rectangular components), by the Law of Sines in the vector triangle:
• F1 / sin β = F2 / sin α = F / sin(α + β).
• Here sin β = sin 45° = 1 / √2, and sin 75° = (√3 + 1) / (2√2).
• Therefore, F1 = 20 × (1 / √2) / [ (√3 + 1) / (2√2) ] = 40 / (√3 + 1) = 20(√3 − 1) N (or if F = 10√2 N, 10√2(√3 − 1) N).
Cross Product of Diagonals (Area of Parallelogram from Diagonals)
JEE Main 2023 (Apr Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q12. The two diagonals of a parallelogram are given by the vectors d1 = (3 î + ĵ − 2 k̂) m and d2 = (î − 3 ĵ + 4 k̂) m. What is the area of the parallelogram?
✔ Half-Cross-Product of Diagonals Area = (1/2)|d1 × d2| Mastered! +4 Marks
✖ When DIAGONALS d1 and d2 are given (instead of adjacent sides), the area of the parallelogram is (1/2)|d1 × d2|!
Concept & Solution:• Since d1 = A + B and d2 = A − B, we have d1 × d2 = −2 (A × B).
• Thus, Area of Parallelogram = (1 / 2) |d1 × d2|.
• Computing d1 × d2: î(4 − 6) − ĵ(12 − (−2)) + k̂(−9 − 1) = −2 î − 14 ĵ − 10 k̂.
• |d1 × d2| = √(4 + 196 + 100) = √300 = 10√3 ⇒ Area = (1 / 2)(10√3) = 5√3 m2.
Vector Addition (Right-Triangle Sides 3, 4, 5 Angle Between Two Vectors)
JEE Main 2022 (July Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q13. Three vectors A, B, and C satisfy the relation A + B = C, and their magnitudes are |A| = 12, |B| = 5, and |C| = 13. What is the angle between vectors A and B, and what is the angle between vectors A and C?
✔ Pythagorean Triplet (122 + 52 = 132) Right-Triangle Geometry Mastered! +4 Marks
✖ Since 122 + 52 = 144 + 25 = 169 = 132, A and B are perpendicular (90°), and cos(∠A,C) = |A|/|C| = 12/13!
Concept & Solution:• Since |A|2 + |B|2 = 122 + 52 = 144 + 25 = 169 = 132 = |C|2:
• Comparing with |C|2 = |A|2 + |B|2 + 2|A||B| cos θ gives cos θ = 0 ⇒ θAB = 90°.
• In this right-angled vector triangle with hypotenuse |C| = 13, the angle α between A and C satisfies cos α = |A| / |C| = 12 / 13.
Time-Dependent Orthogonal Vectors (A(t) · B(t) = 0)
JEE Main 2025 (Apr Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q14. Two time-dependent vectors are given by A(t) = cos(ω t) î + sin(ω t) ĵ and B(t) = cos(ω t / 2) î + sin(ω t / 2) ĵ. At what earliest positive instant of time t are the two vectors A(t) and B(t) mutually perpendicular (orthogonal) to each other?
✔ Cosine Difference Identity A · B = cos(ωt − ωt/2) = 0 Mastered! +4 Marks
✖ A · B = cos(ωt)cos(ωt/2) + sin(ωt)sin(ωt/2) = cos(ωt/2) = 0 ⇒ ωt/2 = π/2 ⇒ t = π / ω!
Concept & Solution:• For orthogonality, set the dot product A(t) · B(t) = 0:
• cos(ω t) cos(ω t / 2) + sin(ω t) sin(ω t / 2) = 0.
• Using cos(α − β) = cos α cos β + sin α sin β:
• cos(ω t − ω t / 2) = cos(ω t / 2) = 0.
• For the earliest positive time: ω t / 2 = π / 2 ⇒ t = π / ω.
Set 30 Grand Finale (450-PYQ Milestone: Lami’s Minimum Force on Inclined Plane)
JEE Main 2026 (Apr Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q15. A smooth block of weight W = 100 N rests on a frictionless inclined plane of inclination α = 30° with the horizontal. It is held in static equilibrium by an external force F applied at an angle θ with the inclined plane. By Lami’s Theorem, for what value of θ is the required force F minimum, and what are that minimum force (Fmin) and the corresponding normal reaction (N) from the incline?
✔ 450th Question Mega-Milestone Conquered! F = W sin α / cos θ Minimized at θ = 0°! +4 Marks
✖ Resolving along the smooth incline gives F cos θ = W sin 30° = 50 N, so F = 50 / cos θ is minimum when cos θ = 1 (θ = 0°, parallel to the incline)!
Concept & Solution:• By Lami’s Theorem (or resolving along the frictionless incline):
• F cos θ = W sin α ⇒ F = (W sin 30°) / cos θ = 50 / cos θ.
• Clearly, F is minimum when cos θ = 1 ⇒ θ = 0° (force directed up and parallel to the inclined plane), giving Fmin = 50 N.
• When θ = 0°, the normal reaction is simply N = W cos 30° = 100 × (√3 / 2) = 50√3 N.
450-PYQ Mega Milestone Completed!
Congratulations! With Set 30 complete, you now have an extraordinary 450-question, zero-dependency JEE Main PYQ bank across Sets 1–30.
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