TOPIC: Scalars vs Vectors, Addition, Resolution, Dot/Cross Products, and Lami’s Theorem
IIT-JEE MAINS PYQ Arena — Set 32
Scalars vs Vectors, Addition, Resolution, Dot/Cross Products & Lami’s Theorem
Set 32 • 480-PYQ Series
Dot Product (Scalar Projection of A on B)
JEE Main 2024 (Jan Shift 1)
⏱ JEE Standard Pace: 120s
+4 / -1 Scale
Q1. What is the scalar projection of the vector A = 2 î + 3 ĵ + 2 k̂ on the vector B = î + 2 ĵ + 2 k̂?
✔ Simple Projection (A · B) / |B| Nailed! +4 Marks
✖ The projection of A on B is given by (A · B) / |B|. Here A · B = 12 and |B| = 3, so 12/3 = 4!
Concept & Solution:• Scalar projection of A on B = (A · B) / |B|.
• Dot product: A · B = (2)(1) + (3)(2) + (2)(2) = 2 + 6 + 4 = 12.
• Magnitude of B: |B| = √(12 + 22 + 22) = √9 = 3.
• Projection = 12 / 3 = 4.
Question 1 of 15
Lami’s Theorem (Symmetric Suspension at 45°)
JEE Main 2025 (Apr Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q2. A mass of 10√2 kg is suspended in static equilibrium by two identical strings that each make an angle of 45° with the horizontal ceiling. Taking g = 10 m s−2, what is the tension T in each string?
✔ Vertical Component Balance 2 T sin 45° = W Mastered! +4 Marks
✖ The upward force is 2 T sin 45° = 2 T / √2 = T√2. Setting this equal to W = 100√2 gives T = 100 N!
Concept & Solution:• Weight of the mass: W = m g = 10√2 × 10 = 100√2 N.
• In equilibrium, the sum of the vertical components of tension balances the weight.
• 2 T sin 45° = W ⇒ 2 T (1 / √2) = 100√2.
• T √2 = 100√2 ⇒ T = 100 N.
Cross Product Application (Area of Triangle)
JEE Main 2024 (Apr Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q3. A triangle is formed by three vertices located at the origin O(0, 0) and the points given by the position vectors a = 3 î + 4 ĵ and b = −3 î + 4 ĵ. What is the area of this triangle?
✔ Triangle Area = (1/2)|a × b| Calculated Correctly! +4 Marks
✖ Remember the (1/2) factor for a triangle! a × b = (3(4) − 4(−3))k̂ = 24k̂. Half of magnitude is 12!
Concept & Solution:• The area of a triangle formed by vectors a and b from the origin is Area = (1 / 2) |a × b|.
• a × b = (3 î + 4 ĵ) × (−3 î + 4 ĵ) = 12(î × ĵ) − 12(ĵ × î) = 12 k̂ − 12(−k̂) = 24 k̂.
• Magnitude |a × b| = 24.
• Area = (1 / 2) × 24 = 12 sq units.
Vector Addition (Equality of Sum and Difference Magnitudes)
JEE Main 2023 (Jan Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q4. If A and B are two non-zero vectors such that the magnitude of their sum is equal to the magnitude of their difference (|A + B| = |A − B|), what must be the angle θ between the two vectors?
✔ Orthogonality Condition Derived! +4 Marks
✖ Square both sides: A² + B² + 2AB cos θ = A² + B² − 2AB cos θ ⇒ 4AB cos θ = 0 ⇒ θ = 90°!
Concept & Solution:• Squaring both sides of |A + B| = |A − B|:
• |A|2 + |B|2 + 2 |A| |B| cos θ = |A|2 + |B|2 − 2 |A| |B| cos θ.
• 4 |A| |B| cos θ = 0 ⇒ cos θ = 0 (since |A| and |B| are non-zero).
• Therefore, θ = 90° (the vectors form the diagonals of a rectangle, which are equal only if it is a rectangle).
Cross & Dot Product (Lagrange’s Identity Application)
JEE Main 2026 (Jan Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q5. For two vectors A and B, it is given that the magnitude of vector A is 4, and the magnitudes of their cross product and dot product are related by |A × B|2 + |A · B|2 = 144. What is the magnitude of vector B?
✔ Lagrange’s Identity |A|²|B|² = 144 Mastered! +4 Marks
✖ Use Lagrange’s identity: |A × B|2 + (A · B)2 = |A|2 |B|2. So 16 |B|2 = 144 ⇒ |B| = 3!
Concept & Solution:• According to Lagrange’s vector identity: |A × B|2 + (A · B)2 = |A|2 |B|2.
• Substituting the given values: 144 = (4)2 × |B|2 = 16 |B|2.
• |B|2 = 144 / 16 = 9 ⇒ |B| = 3.
Scalar Triple Product (Coplanar Vectors Parameter α)
JEE Main 2022 (July Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q6. If the three vectors a = î + ĵ + k̂, b = î − ĵ + k̂, and c = î + 2 ĵ + α k̂ are coplanar, what is the value of the constant α?
✔ Determinant [a b c] = 0 Evaluated Correctly! +4 Marks
✖ Calculate a × b = 2î − 2k̂. For coplanarity, c · (a × b) = 0 ⇒ (1)(2) + (2)(0) + (α)(−2) = 0 ⇒ 2 − 2α = 0 ⇒ α = 1!
Concept & Solution:• For three vectors to be coplanar, their scalar triple product [a b c] = a · (b × c) (or determinant of components) must be zero.
• First compute a × b: î(−1 − (−1)) − ĵ(1 − 1) + k̂(−1 − 1) = 0 î − 0 ĵ − 2 k̂? Wait.
• Let’s re-calculate: a × b = î(1 − (−1)) − ĵ(1 − 1) + k̂(−1 − 1) = 2 î − 2 k̂.
• Now, c · (a × b) = 0 ⇒ (1)(2) + (2)(0) + (α)(−2) = 0 ⇒ 2 − 2α = 0 ⇒ α = 1.
Dot Product Orthogonal Condition Expansion
JEE Main 2023 (Apr Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q7. Two non-zero vectors A and B have equal magnitudes (|A| = |B|). If the composite vector (2 A + B) is perpendicular to the composite vector (A − 2 B), what is the angle θ between vectors A and B?
✔ Dot Product Expansion (2A² − 3A·B − 2B² = 0) Mastered! +4 Marks
✖ Expand the dot product: (2A + B) · (A − 2B) = 2A² − 4A·B + B·A − 2B² = 0. Since A² = B², this leaves −3A·B = 0 ⇒ θ = 90°!
Concept & Solution:• Orthogonality condition: (2 A + B) · (A − 2 B) = 0.
• Expanding the dot product: 2(A · A) − 4(A · B) + (B · A) − 2(B · B) = 0.
• 2|A|2 − 3(A · B) − 2|B|2 = 0.
• Since |A| = |B|, the squared terms cancel: 2|A|2 − 2|B|2 = 0.
• This leaves −3 (A · B) = 0 ⇒ A · B = 0, meaning the vectors are perpendicular (θ = 90°).
Lami’s Theorem (Force Ratio for Three Coplanar Forces)
JEE Main 2024 (Jan Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q8. Three coplanar forces F1, F2, and F3 keep a body in static equilibrium. The angle between F1 and F2 is 90°, and the angle between F2 and F3 is 150°. What is the ratio of their magnitudes F1 : F2 : F3?
✔ Lami’s Ratio F ∝ sin(opposite angle) Mastered! +4 Marks
✖ The angle opposite F1 is 150°, opposite F2 is 360°−(90°+150°) = 120°, and opposite F3 is 90°. F1:F2:F3 = sin 150° : sin 120° : sin 90° = 1/2 : √3/2 : 1 = 1 : √3 : 2!
Concept & Solution:• The three angles between the forces are 90°, 150°, and 360° − (90° + 150°) = 120°.
• Angle opposite F1 is 150°; opposite F2 is 120°; opposite F3 is 90°.
• By Lami’s Theorem: F1 / sin 150° = F2 / sin 120° = F3 / sin 90°.
• Ratio = sin 150° : sin 120° : sin 90° = (1/2) : (√3/2) : 1.
• Multiplying by 2 yields 1 : √3 : 2.
Unit Vector Parallel to Resultant
JEE Main 2021 (Aug Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q9. What is the unit vector parallel to the resultant of the two vectors A = 4 î − 3 ĵ and B = 8 î + 8 ĵ?
✔ Vector Addition & Normalization R/|R| Nailed! +4 Marks
✖ R = A + B = 12î + 5ĵ. Magnitude |R| = √(144 + 25) = 13. Divide R by 13 to get the unit vector!
Concept & Solution:• Resultant vector R = A + B = (4 + 8)î + (−3 + 8)ĵ = 12 î + 5 ĵ.
• Magnitude of Resultant |R| = √(122 + 52) = √(144 + 25) = √169 = 13.
• Unit vector = R / |R| = (12 î + 5 ĵ) / 13.
Resolution of Vectors (Perpendicular Component Derivation)
JEE Main 2023 (Apr Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q10. Given two vectors A = 3 î + 4 ĵ and B = î + ĵ, what is the vector component of A that is strictly perpendicular to B (in the same plane)?
✔ Vector Subtraction A⊥ = A − A∥ Mastered! +4 Marks
✖ A∥ = [(A·B)/|B|²] B = (7/2)(î + ĵ) = 3.5î + 3.5ĵ. Subtract this from A to get A⊥ = (3−3.5)î + (4−3.5)ĵ = −0.5î + 0.5ĵ!
Concept & Solution:• Parallel component: A∥ = [ (A · B) / |B|2 ] B = [ (3 + 4) / 2 ] (î + ĵ) = 3.5 î + 3.5 ĵ.
• Perpendicular component: A⊥ = A − A∥ = (3 î + 4 ĵ) − (3.5 î + 3.5 ĵ) = −0.5 î + 0.5 ĵ.
• (Check orthogonality: (−0.5 î + 0.5 ĵ) · (î + ĵ) = −0.5 + 0.5 = 0 ✔).
Cross Product (Torque Vector τ = r × F)
JEE Main 2025 (Jan Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q11. A force F = î + 2 ĵ − k̂ N acts on a particle located at position vector r = 2 î − ĵ + 3 k̂ m. What is the torque vector τ acting on the particle about the origin?
✔ Right-Hand Rule Order (τ = r × F) Mastered! +4 Marks
✖ Torque must be calculated as r × F, NOT F × r! Reversing the order flips all the signs!
Concept & Solution:• Torque τ = r × F.
• τ = î [ (−1)(−1) − (3)(2) ] − ĵ [ (2)(−1) − (3)(1) ] + k̂ [ (2)(2) − (−1)(1) ]
• τ = î (1 − 6) − ĵ (−2 − 3) + k̂ (4 + 1)
• τ = −5 î + 5 ĵ + 5 k̂ N·m.
Vector Addition (Minimum Coplanar Unequal Vectors for Equilibrium)
JEE Main 2022 (July Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q12. What is the minimum number of coplanar vectors of unequal magnitudes required such that their vector sum can be zero?
✔ Triangle Closure Property Mastered! +4 Marks
✖ Two unequal vectors can never sum to zero! You need a minimum of 3 unequal vectors to form a closed triangle!
Concept & Solution:• To have a resultant of zero with exactly 2 vectors, they must be strictly equal in magnitude and opposite in direction.
• If all magnitudes are unequal, you cannot close the polygon with just 2 vectors.
• You need a minimum of 3 unequal vectors (which form a scalene triangle satisfying the triangle inequality) for their sum to be zero.
Vector Addition (Resultant Perpendicular Condition)
JEE Main 2026 (Apr Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q13. The resultant R of two vectors P and Q is perpendicular to P, and its magnitude is exactly half of the magnitude of Q (i.e., R = Q / 2). What is the angle θ between the vectors P and Q?
✔ Right-Triangle Velocity Closure R = Q sin(180−θ) Nailed! +4 Marks
✖ In the vector triangle, R is the opposite side to angle (180°−θ) with hypotenuse Q. So R = Q sin(180°−θ) = Q sin θ. Since R = Q/2, sin θ = 1/2. Since the angle must be obtuse, θ = 150°!
Concept & Solution:• Since R is perpendicular to P, P, R, and Q form a right-angled triangle with Q as the hypotenuse.
• From vector addition geometry, R = Q sin(180° − θ) = Q sin θ.
• We are given R = Q / 2 ⇒ Q sin θ = Q / 2 ⇒ sin θ = 1 / 2.
• Since R is perpendicular to P, the angle between P and Q must be obtuse: θ = 150°.
Dot Product (Sine of Angle Between Two Vectors)
JEE Main 2024 (Jan Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q14. Given two vectors A = 2 î + ĵ + 2 k̂ and B = 2 î − 2 ĵ + k̂, what is the value of sin θ, where θ is the angle between them?
✔ Dot Product to Cosine to Sine Conversion Mastered! +4 Marks
✖ A·B = 4, |A| = 3, |B| = 3 ⇒ cos θ = 4/9. Then sin θ = √(1 − cos²θ) = √(1 − 16/81) = √65 / 9!
Concept & Solution:• Dot Product: A · B = (2)(2) + (1)(−2) + (2)(1) = 4 − 2 + 2 = 4.
• Magnitudes: |A| = √(4+1+4) = 3 ; |B| = √(4+4+1) = 3.
• cos θ = (A · B) / (|A| |B|) = 4 / 9.
• Using sin2θ + cos2θ = 1: sin θ = √(1 − (16 / 81)) = √(65 / 81) = √65 / 9.
Set 32 Grand Finale (Scalar Triple Product of Repeated Vectors)
JEE Main 2023 (Apr Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q15. For any two arbitrary 3D vectors A and B, what is the value of the scalar triple product expression A · (A × B)?
✔ 480th Question Milestone Conquered! Orthogonal Dot Product = 0! +4 Marks
✖ The vector (A × B) is strictly perpendicular to A. The dot product of two perpendicular vectors is always ZERO!
Concept & Solution:• By definition of the cross product, the vector (A × B) is perpendicular to the plane containing both A and B.
• Therefore, (A × B) is strictly perpendicular to A (and to B).
• The dot product of two perpendicular vectors is always zero: A · (A × B) = 0.
Set 32 Vector Ops Series Completed!
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