TOPIC: Scalars vs Vectors, Addition, Resolution, Dot/Cross Products, and Lami’s Theorem
IIT-JEE MAINS PYQ Arena — Set 31
Scalars vs Vectors, Addition, Resolution, Dot/Cross Products & Lami’s Theorem
Set 31 • Vectors Pro
Vector Addition (Max & Min Resultant Magnitudes)
JEE Main 2021 (Aug Shift 1)
⏱ JEE Standard Pace: 120s
+4 / -1 Scale
Q1. The maximum and minimum magnitudes of the resultant of two vectors P and Q are 17 units and 7 units respectively. If these two vectors act at right angles (90°) to each other, what will be the magnitude of their resultant?
✔ Simple Linear Equations + Pythagorean Theorem! +4 Marks
✖ Rmax = P+Q = 17 and Rmin = P-Q = 7. Thus P = 12, Q = 5. At 90°, R = √(12² + 5²) = 13!
Concept & Solution:• Maximum resultant occurs at θ = 0°: P + Q = 17.
• Minimum resultant occurs at θ = 180°: P − Q = 7 (assuming P > Q).
• Adding the equations: 2P = 24 ⇒ P = 12 units, and Q = 5 units.
• When θ = 90°, Resultant R = √(P2 + Q2) = √(122 + 52) = √(144 + 25) = √169 = 13 units.
Question 1 of 15
Dot Product (Scalar Projection)
JEE Main 2022 (July Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q2. Given two vectors A = î + ĵ + k̂ and B = 2 î + 2 ĵ + k̂, what is the magnitude of the scalar projection of vector A on vector B?
✔ Dot Product / Magnitude of Base Vector Mastered! +4 Marks
✖ Scalar projection is A · B / |B|. A · B = 2 + 2 + 1 = 5, and |B| = √(4 + 4 + 1) = 3!
Concept & Solution:• The scalar projection of A on B is given by (A · B) / |B|.
• Dot Product: A · B = (1)(2) + (1)(2) + (1)(1) = 2 + 2 + 1 = 5.
• Magnitude of B: |B| = √(22 + 22 + 12) = √9 = 3.
• Projection = 5 / 3.
Vector Identity: |a × b|² + (a · b)²
JEE Main 2021 (Mar Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q3. For two non-zero vectors a and b, it is given that |a| = 2, |b| = 5, and their scalar product a · b = 8. What is the magnitude of their cross product, |a × b|?
✔ Lagrange’s Identity Applied Directly! +4 Marks
✖ |a × b|2 + (a · b)2 = |a|2 |b|2 ⇒ x2 + 64 = 100 ⇒ x = 6!
Concept & Solution:• According to Lagrange’s vector identity: |a × b|2 + (a · b)2 = |a|2 |b|2.
• Substituting the known values: |a × b|2 + (8)2 = (2)2 × (5)2.
• |a × b|2 + 64 = 4 × 25 = 100.
• |a × b|2 = 100 − 64 = 36 ⇒ |a × b| = 6.
Area of Parallelogram via Cross Product
JEE Main 2022 (Aug Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q4. Let two adjacent sides of a parallelogram be given by vectors a = î + α ĵ + 3 k̂ and b = 3 î − α ĵ + k̂. If the area of the parallelogram formed by them is 8√3 square units, what is the positive value of α?
✔ Cross Product Determinant Magnitude Solved Algebraically! +4 Marks
✖ a × b = 4α î + 8 ĵ − 4α k̂. Setting |a × b|2 = 192 gives 32α2 + 64 = 192 ⇒ α = 2!
Concept & Solution:• Area = |a × b|. First, compute a × b:
• î [α(1) − 3(−α)] − ĵ [1(1) − 3(3)] + k̂ [1(−α) − 3(α)]
• = 4α î + 8 ĵ − 4α k̂.
• |a × b|2 = (4α)2 + 82 + (−4α)2 = 16α2 + 64 + 16α2 = 32α2 + 64.
• Area = 8√3 ⇒ Area2 = 64 × 3 = 192.
• 32α2 + 64 = 192 ⇒ 32α2 = 128 ⇒ α2 = 4 ⇒ α = 2.
Resultant of Forces Modifying Under Scaling
JEE Main 2025 (Apr Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q5. The resultant of two forces 3P and 2P is R. If the first force is doubled (becomes 6P), the magnitude of the resultant is also doubled (becomes 2R). What is the angle θ between the two forces?
✔ Vector Addition Substitution Mastered! +4 Marks
✖ R2 = 13P2 + 12P2 cos θ, and 4R2 = 40P2 + 24P2 cos θ. Substitute R2 to get cos θ = −1/2!
Concept & Solution:• Case 1: R2 = (3P)2 + (2P)2 + 2(3P)(2P) cos θ = 13P2 + 12P2 cos θ.
• Case 2: (2R)2 = (6P)2 + (2P)2 + 2(6P)(2P) cos θ ⇒ 4R2 = 40P2 + 24P2 cos θ.
• Substitute R2 into Case 2: 4(13P2 + 12P2 cos θ) = 40P2 + 24P2 cos θ.
• 52P2 + 48P2 cos θ = 40P2 + 24P2 cos θ ⇒ 24 cos θ = −12 ⇒ cos θ = −1/2 ⇒ θ = 120°.
Equilibrium (Resolution of Tensions)
JEE Main 2023 (Jan Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q6. A block of mass 10 kg is suspended by two light strings. String 1 is strictly horizontal, while String 2 makes an angle of 30° with the vertical. The block hangs in static equilibrium. Taking g = 10 m s−2, what is the tension in String 2?
✔ T2 cos 30° = mg Solved Flawlessly! +4 Marks
✖ The only vertical force to balance weight mg is the vertical component of T2: T2 cos 30° = 100 ⇒ T2 = 200 / √3!
Concept & Solution:• Let T2 be the tension in the string at 30° to the vertical.
• The vertical component of T2 balances the block’s weight: T2 cos 30° = m g = 100 N.
• T2 (√3 / 2) = 100 ⇒ T2 = 200 / √3 N.
• (The horizontal string’s tension would be T1 = T2 sin 30° = 100 / √3 N).
Scalar Triple Product (Coplanarity Condition)
JEE Main 2022 (July Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q7. The three vectors A = î + ĵ + k̂, B = î − ĵ + k̂, and C = î + 2 ĵ + α k̂ are coplanar. What must be the value of α?
✔ Determinant [A B C] = 0 Mastered! +4 Marks
✖ Evaluate the determinant of the coefficients: 1(−α − 2) − 1(α − 1) + 1(2 − (−1)) = 0 ⇒ −2α + 2 = 0 ⇒ α = 1!
Concept & Solution:• For three vectors to be coplanar, their scalar triple product (determinant) must be zero.
• Det = 1( (−1)(α) − (1)(2) ) − 1( (1)(α) − (1)(1) ) + 1( (1)(2) − (−1)(1) ) = 0.
• (−α − 2) − (α − 1) + (2 + 1) = 0.
• −2α − 2 + 1 + 3 = 0 ⇒ −2α + 2 = 0 ⇒ α = 1.
Angle Between Sum and Difference of Equal Vectors
JEE Main 2025 (Jan Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q8. If A and B are two non-zero vectors such that their magnitudes are equal (|A| = |B|), what is the angle between the sum vector (A + B) and the difference vector (A − B)?
✔ Dot Product (A+B) · (A−B) = A² − B² = 0 Nailed! +4 Marks
✖ Since |A| = |B|, their dot product (A + B) · (A − B) = |A|2 − |B|2 = 0, meaning they are strictly perpendicular (90°)!
Concept & Solution:• Let’s take the dot product of the two vectors: (A + B) · (A − B).
• Expanding gives: A · A − A · B + B · A − B · B = |A|2 − |B|2.
• Since |A| = |B|, we get |A|2 − |B|2 = 0.
• A dot product of zero implies the two vectors are strictly orthogonal: Angle = 90° (diagonals of a rhombus are always perpendicular).
Unit Vector Normal to Two Vectors (Cross Product)
JEE Main 2023 (Apr Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q9. What is the unit vector perpendicular to both vectors A = 2 î − ĵ + k̂ and B = 3 î + 4 ĵ − k̂?
✔ Cross Product n̂ = (A × B) / |A × B| Solved Properly! +4 Marks
✖ The cross product A × B = î(1 − 4) − ĵ(−2 − 3) + k̂(8 − (−3)) = −3î + 5ĵ + 11k̂. Divide by magnitude √155!
Concept & Solution:• A × B = î [ (−1)(−1) − (1)(4) ] − ĵ [ (2)(−1) − (1)(3) ] + k̂ [ (2)(4) − (−1)(3) ]
• = î (1 − 4) − ĵ (−2 − 3) + k̂ (8 + 3) = −3 î + 5 ĵ + 11 k̂.
• Magnitude = √((−3)2 + 52 + 112) = √(9 + 25 + 121) = √155.
• Unit vector = (−3 î + 5 ĵ + 11 k̂) / √155.
Coplanar Force Polygon Vector Summation
JEE Main 2024 (Jan Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q10. N coplanar forces of equal magnitude F act simultaneously on a particle. The angle between any two adjacent forces is identically 2π / N (forming a symmetric radial star pattern). If one of these forces is suddenly reversed in direction, what is the magnitude of the new net resultant force acting on the particle?
✔ Symmetric Force Cancellation Logic Mastered! +4 Marks
✖ Reversing F₁ is equivalent to removing F₁ and then applying −F₁! Initial sum is 0. Removing F₁ gives resultant −F₁, then adding reversed F₁ gives −2F₁!
Concept & Solution:• Initially, the N symmetric forces form a closed regular polygon, so their net sum is zero: ∑ Fi = F1 + F2 + … + FN = 0.
• Let force F1 be reversed. The new sum is: (−F1) + F2 + … + FN.
• We can rewrite this as: (F1 + F2 + … + FN) − 2 F1 = 0 − 2 F1 = −2 F1.
• The magnitude of the new net force is |−2 F1| = 2 F.
3D Direction Cosines Constraint
JEE Main 2023 (Jan Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q11. A 3D vector makes an angle of 45° with the positive x-axis and an angle of 60° with the positive y-axis. What is the acute angle γ it makes with the positive z-axis?
✔ Direction Cosine Sum cos²α + cos²β + cos²γ = 1 Mastered! +4 Marks
✖ cos²(45°) + cos²(60°) + cos²γ = 1 ⇒ 1/2 + 1/4 + cos²γ = 1 ⇒ cos²γ = 1/4 ⇒ cos γ = 1/2 ⇒ γ = 60°!
Concept & Solution:• Direction cosines must satisfy: cos2α + cos2β + cos2γ = 1.
• Given α = 45° and β = 60°:
• (1 / √2)2 + (1 / 2)2 + cos2γ = 1 ⇒ 1 / 2 + 1 / 4 + cos2γ = 1.
• 3 / 4 + cos2γ = 1 ⇒ cos2γ = 1 / 4 ⇒ cos γ = 1 / 2 (for acute angle).
• Therefore, γ = 60°.
Cross Product Area (Parallelogram from Diagonals)
JEE Main 2022 (July Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q12. The two diagonals of a parallelogram are represented by the vectors d1 = 3 î − 4 ĵ and d2 = 2 î + 3 k̂. What is the total area of the parallelogram?
✔ Area from Diagonals = (1/2) |d₁ × d₂| Mastered! +4 Marks
✖ When diagonals are given, Area = (1/2)|d1 × d2|! d1 × d2 = −12î − 9ĵ + 8k̂. Magnitude = 17, so Area = 17/2 = 8.5!
Concept & Solution:• For a parallelogram defined by its diagonals d1 and d2, Area = (1 / 2) |d1 × d2|.
• Cross Product: d1 × d2 = (3 î − 4 ĵ) × (2 î + 3 k̂)
• = 6(î × î) + 9(î × k̂) − 8(ĵ × î) − 12(ĵ × k̂)
• = 0 − 9ĵ + 8k̂ − 12î = −12 î − 9 ĵ + 8 k̂.
• Magnitude = √((−12)2 + (−9)2 + 82) = √(144 + 81 + 64) = √289 = 17.
• Area = 17 / 2 = 8.5 sq. units.
Dot Product Application (Work Done by Force)
JEE Main 2024 (Jan Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q13. A constant force F = (4 î + ĵ + 3 k̂) N acts on a particle and displaces it from position A(1, 1, 1) m to position B(3, 3, 3) m. What is the total work done (W = F · Δr) by the force on the particle?
✔ Displacement Δr = r₂ − r₁ & Dot Product Nailed! +4 Marks
✖ Δr = (3−1)î + (3−1)ĵ + (3−1)k̂ = 2î + 2ĵ + 2k̂ m. W = F · Δr = 4(2) + 1(2) + 3(2) = 8 + 2 + 6 = 16 J!
Concept & Solution:• Displacement vector: Δr = rB − rA = (3 − 1)î + (3 − 1)ĵ + (3 − 1)k̂ = (2 î + 2 ĵ + 2 k̂) m.
• Work Done W = F · Δr = (4 î + ĵ + 3 k̂) · (2 î + 2 ĵ + 2 k̂).
• W = (4 × 2) + (1 × 2) + (3 × 2) = 8 + 2 + 6 = 16 J.
Cross Product Application (Angular Momentum L = r × p)
JEE Main 2025 (Apr Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q14. The position vector of a particle is r = (2 î − ĵ + k̂) m and its linear momentum is p = (î + 2 ĵ − k̂) kg m s−1. What is the angular momentum (L = r × p) of the particle about the origin?
✔ Determinant Expansion L = r × p Executed Perfectly! +4 Marks
✖ Evaluate r × p: î(1 − 2) − ĵ(−2 − 1) + k̂(4 − (−1)) = −î + 3ĵ + 5k̂!
Concept & Solution:• L = r × p.
• L = î [ (−1)(−1) − (1)(2) ] − ĵ [ (2)(−1) − (1)(1) ] + k̂ [ (2)(2) − (−1)(1) ]
• L = î (1 − 2) − ĵ (−2 − 1) + k̂ (4 + 1)
• L = −î + 3 ĵ + 5 k̂ (kg m2 s−1).
Set 31 Grand Finale (Change in Velocity Vector Δv)
JEE Main 2026 (Apr Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q15. A particle initially moves with a velocity vector v1 = (3 î + 4 ĵ) m s−1. Due to the action of a constant force, its velocity changes to v2 = (−3 î + 4 ĵ) m s−1. What is the magnitude of the change in velocity vector (|Δv|)?
✔ Vector Subtraction Δv = v₂ − v₁ Nailed! +4 Marks
✖ Do not just subtract the magnitudes (5 − 5 = 0)! Δv = v2 − v1 = (−3î + 4ĵ) − (3î + 4ĵ) = −6î, so magnitude is 6 m s−1!
Concept & Solution:• Change in velocity vector: Δv = v2 − v1.
• Δv = (−3 î + 4 ĵ) − (3 î + 4 ĵ) = −6 î + 0 ĵ.
• Magnitude: |Δv| = √( (−6)2 + 0 ) = 6 m s−1.
Set 31 Vector Ops Series Completed!
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