TOPIC: Scalars vs Vectors, Addition, Resolution, Dot/Cross Products, and Lami’s Theorem
IIT-JEE MAINS PYQ Arena — Set 29
Scalars vs Vectors, Addition, Resolution, Dot/Cross Products & Lami’s Theorem
Set 29 • 435-PYQ Series
Vector Addition & Subtraction (|A + B| = n |A − B|)
JEE Main 2024 (Jan Shift 1)
⏱ JEE Standard Pace: 120s
+4 / -1 Scale
Q1. Two vectors A and B have equal magnitudes (|A| = |B| = F). If the magnitude of their sum |A + B| is √3 times the magnitude of their difference |A − B|, what is the angle θ between vectors A and B?
✔ Half-Angle Shortcut cot(θ/2) = √3 ⇒ θ = 60° Mastered! +4 Marks
✖ For two equal vectors, |A + B| = 2F cos(θ/2) and |A − B| = 2F sin(θ/2), so tan(θ/2) = 1/√3 ⇒ θ/2 = 30° ⇒ θ = 60°!
Concept & Solution:• For two vectors of equal magnitude F: |A + B| = 2 F cos(θ / 2) and |A − B| = 2 F sin(θ / 2).
• Given |A + B| = √3 |A − B| ⇒ 2 F cos(θ / 2) = √3 [ 2 F sin(θ / 2) ].
• Therefore, tan(θ / 2) = 1 / √3 ⇒ θ / 2 = 30° ⇒ θ = 60°.
Question 1 of 15
Resultant Perpendicular to Smaller Vector
JEE Main 2024 (Apr Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q2. The sum of the magnitudes of two forces acting at a point is 18 N, and the magnitude of their resultant is 12 N. If the resultant force is at an angle of 90° (perpendicular) to the force of smaller magnitude, what are the magnitudes of the two individual forces?
✔ Right-Triangle Vector Closure F22 − F12 = R2 Mastered! +4 Marks
✖ Since R ⊥ F1, F2 is the hypotenuse: F22 − F12 = 122 = 144! Dividing by F1 + F2 = 18 gives F2 − F1 = 8 N!
Concept & Solution:• Given: F1 + F2 = 18 N and resultant R = 12 N is perpendicular to smaller force F1.
• By head-to-tail triangle closure, F1, R, and F2 form a right triangle with F2 as the hypotenuse:
• F22 − F12 = R2 = 122 = 144.
• Since (F2 − F1)(F2 + F1) = 144 ⇒ F2 − F1 = 144 / 18 = 8 N.
• Solving F2 + F1 = 18 and F2 − F1 = 8 gives F1 = 5 N and F2 = 13 N.
Lami’s Theorem (Three Concurrent Forces in Equilibrium)
JEE Main 2025 (Jan Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q3. A block of weight W = 100√3 N is suspended in equilibrium from a rigid ceiling by two light inextensible strings OA and OB making angles of 60° and 30° with the horizontal ceiling respectively, as shown below. Using Lami’s Theorem, what are the tensions T1 (in OA) and T2 (in OB)?
✔ Lami’s Sine Rule T1/sin(120°) = T2/sin(150°) = W/sin(90°) Mastered! +4 Marks
✖ Angle between T1 and T2 is 90°! Angle opposite T1 is 90° + 30° = 120°, and angle opposite T2 is 90° + 60° = 150°!
Concept & Solution:• Angle between the two strings OA and OB is 180° − (60° + 30°) = 90°.
• Angle opposite T1 (between T2 and W) is 90° + 30° = 120°.
• Angle opposite T2 (between T1 and W) is 90° + 60° = 150°.
• By Lami’s Theorem: T1 / sin 120° = T2 / sin 150° = W / sin 90° = 100√3.
• Therefore, T1 = 100√3 × (√3 / 2) = 150 N and T2 = 100√3 × (1 / 2) = 50√3 N.
Dot Product & Orthogonality Condition
JEE Main 2024 (Jan Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q4. If vector A = (2 î + 3 ĵ + 8 k̂) is perpendicular to vector B = (4 ĵ − 4 î + α k̂), then what is the value of α?
✔ Swapped î and ĵ Order in Vector B Caught! +4 Marks
✖ Trap Alert! Look closely at the unit vectors in B: it is written as (4 ĵ − 4 î + α k̂), which is (−4 î + 4 ĵ + α k̂)!
Concept & Solution:• Rewriting vector B in standard (î, ĵ, k̂) order: B = −4 î + 4 ĵ + α k̂.
• For perpendicular vectors, their scalar (dot) product must be zero: A · B = 0.
• (2)(−4) + (3)(4) + (8)(α) = 0 ⇒ −8 + 12 + 8 α = 0 ⇒ 8 α = −4 ⇒ α = −1 / 2.
Cross Product vs. Dot Product Relation
JEE Main 2023 (Apr Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q5. Two vectors A and B satisfy the relation |A × B| = √3 (A · B). What is the angle θ between A and B, and what is the magnitude of their resultant |A + B| in terms of A and B?
✔ tan θ = √3 ⇒ θ = 60° & cos 60° = 1/2 Mastered! +4 Marks
✖ AB sin θ = √3 AB cos θ ⇒ tan θ = √3 ⇒ θ = 60°! Plugging cos 60° = 1/2 into √(A2 + B2 + 2AB cos θ) gives (A2 + B2 + AB)1/2!
Concept & Solution:• Since |A × B| = A B sin θ and A · B = A B cos θ:
• A B sin θ = √3 A B cos θ ⇒ tan θ = √3 ⇒ θ = 60°.
• Magnitude of resultant: |A + B| = (A2 + B2 + 2 A B cos 60°)1/2.
• Since cos 60° = 1 / 2, we get |A + B| = (A2 + B2 + A B)1/2.
Resolution of Vectors & Vector Projection
JEE Main 2025 (Apr Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q6. Given two vectors A = (3 î + 4 ĵ) and B = (î + ĵ), what are the scalar projection (component) of vector A along vector B, and the vector component of A parallel to B?
✔ Scalar Projection (A · B̂) vs. Vector Projection (A · B̂) B̂ Mastered! +4 Marks
✖ Scalar projection is A · B / |B| = (3 + 4)/√2 = 7/√2! Multiplying by the unit vector B̂ = (î + ĵ)/√2 gives (7/2)(î + ĵ)!
Concept & Solution:• Unit vector along B: B̂ = B / |B| = (î + ĵ) / √2.
• Scalar projection of A along B: A∥ = A · B̂ = (3 × 1 + 4 × 1) / √2 = 7 / √2.
• Vector component of A parallel to B: A∥ B̂ = (7 / √2) × [ (î + ĵ) / √2 ] = (7 / 2)(î + ĵ).
Unit Vectors Difference |ê1 − ê2| from Sum
JEE Main 2023 (Jan Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q7. If the sum of two unit vectors â and b̂ is also a unit vector (i.e., |â + b̂| = 1), then what is the angle θ between â and b̂, and what is the magnitude of their difference |â − b̂|?
✔ Identity |â + b̂|2 + |â − b̂|2 = 4 Mastered! +4 Marks
✖ Use the parallelogram identity |â + b̂|2 + |â − b̂|2 = 2(12 + 12) = 4! Since |â + b̂|2 = 1, |â − b̂|2 = 3 ⇒ |â − b̂| = √3!
Concept & Solution:• Since |â + b̂| = 2 cos(θ / 2) = 1 ⇒ cos(θ / 2) = 1 / 2 ⇒ θ / 2 = 60° ⇒ θ = 120°.
• Using the parallelogram law identity: |â + b̂|2 + |â − b̂|2 = 2 (|â|2 + |b̂|2) = 4.
• Substituting |â + b̂| = 1 gives: 1 + |â − b̂|2 = 4 ⇒ |â − b̂| = √3.
3D Direction Cosines & Angles with Coordinate Axes
JEE Main 2024 (Apr Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q8. A vector P in 3D space makes an angle of α = 60° with the positive x-axis and an angle of β = 45° with the positive y-axis. What acute angle γ does vector P make with the positive z-axis, and what is the value of sin2α + sin2β + sin2γ?
✔ Direction Cosines cos2α + cos2β + cos2γ = 1 Mastered! +4 Marks
✖ Use cos2(60°) + cos2(45°) + cos2γ = 1 ⇒ 1/4 + 1/2 + cos2γ = 1 ⇒ cos γ = 1/2 ⇒ γ = 60°, and sin2α + sin2β + sin2γ = 3 − 1 = 2!
Concept & Solution:• Fundamental identity of direction cosines: cos2α + cos2β + cos2γ = 1.
• (1 / 2)2 + (1 / √2)2 + cos2γ = 1 ⇒ 1 / 4 + 1 / 2 + cos2γ = 1 ⇒ cos2γ = 1 / 4 ⇒ γ = 60°.
• Also, sin2α + sin2β + sin2γ = (1 − cos2α) + (1 − cos2β) + (1 − cos2γ) = 3 − 1 = 2.
Cross Product Application (Torque τ = r × F & Parallelogram Area)
JEE Main 2025 (Jan Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q9. Two adjacent sides of a parallelogram are represented by the vectors A = (2 î − 3 ĵ + k̂) m and B = (î + ĵ − 2 k̂) m. What is the vector A × B, and what is the area of the parallelogram?
✔ Determinant Cross Product & Parallelogram Area |A × B| Mastered! +4 Marks
✖ Area of a parallelogram with adjacent sides A and B is |A × B| (whereas triangle area is (1/2)|A × B|)!
Concept & Solution:• Expanding the determinant for A × B:
• î [ (−3)(−2) − (1)(1) ] − ĵ [ (2)(−2) − (1)(1) ] + k̂ [ (2)(1) − (−3)(1) ]
• = î (6 − 1) − ĵ (−4 − 1) + k̂ (2 + 3) = 5 î + 5 ĵ + 5 k̂.
• Area of the parallelogram = |A × B| = √(52 + 52 + 52) = 5√3 m2.
Lami’s Theorem (Pendulum Bob Pulled by Horizontal Force)
JEE Main 2023 (Apr Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q10. A metal sphere of mass m = 2 kg suspended by a light string from a hook is pulled aside by a horizontal force F until the string makes an angle of θ = 30° with the vertical and remains in static equilibrium. Taking g = 10 m s−2, what are the magnitudes of the horizontal force F and the string tension T?
✔ Lami’s Equilibrium F = mg tan 30° & T = mg sec 30° Mastered! +4 Marks
✖ T cos 30° = mg = 20 N ⇒ T = 40 / √3 N, and F = T sin 30° = mg tan 30° = 20 / √3 N!
Concept & Solution:• Weight of sphere: W = m g = 2 × 10 = 20 N.
• By Lami’s Theorem (or resolving T into T cos 30° = W and T sin 30° = F):
• T = W / cos 30° = 20 / (√3 / 2) = 40 / √3 N.
• F = W tan 30° = 20 × (1 / √3) = 20 / √3 N.
Scalars vs. Vectors (Finite vs. Infinitesimal Rotation & Electric Current)
JEE Main 2022 (June Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q11. Why are Electric Current (I) in a wire and Finite Angular Displacement (Δθ) in 3D space classified as scalars (not vectors) despite having both magnitude and a sense of direction?
✔ Fundamental Vector-Law Criterion Nailed! +4 Marks
✖ Having magnitude and direction is NOT enough! A physical quantity is a vector ONLY if it obeys the commutative parallelogram law of vector addition!
Concept & Solution:• At a wire junction, currents add algebraically (I1 + I2) independent of the angle between the wires.
• Similarly, finite 3D rotations do not commute (θ1 + θ2 ≠ θ2 + θ1), whereas infinitesimal rotations (dθ) and angular velocity (ω = dθ/dt) do commute and are true axial vectors.
Vector Addition (Triangle Inequality & Zero Resultant Triplet)
JEE Main 2024 (Jan Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q12. Three coplanar forces of magnitudes F1, F2, and F3 act simultaneously on a particle. Which of the following sets of magnitudes (F1, F2, F3) can NEVER produce a zero resultant (equilibrium), regardless of the angles between them?
✔ Triangle Inequality |F1 − F2| ≤ F3 ≤ (F1 + F2) Mastered! +4 Marks
✖ For three vectors to sum to zero, they must form a closed triangle (so the largest magnitude cannot exceed the sum of the other two: 35 N > 10 N + 20 N)!
Concept & Solution:• The resultant of F1 = 10 N and F2 = 20 N must lie between:
• Rmin = |20 − 10| = 10 N and Rmax = 20 + 10 = 30 N.
• Since 35 N > 30 N, the third force of 35 N can never be balanced by 10 N and 20 N.
Scalar Triple Product (Coplanar Vectors Condition)
JEE Main 2025 (Apr Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q13. If the three vectors A = (2 î − ĵ + k̂), B = (î + 2 ĵ − 3 k̂), and C = (3 î + λ ĵ + 5 k̂) are coplanar (lie in the same plane), then what is the value of λ?
✔ Zero Scalar Triple Product C · (A × B) = 0 Mastered! +4 Marks
✖ For three coplanar vectors, their scalar triple product determinant [A B C] = 0! Since A × B = î − 7ĵ − 5k̂, C · (A × B) = 3 − 7λ − 25 = 0 ⇒ λ = −4! Wait: let’s check A × B!
Concept & Solution:• First compute A × B: î(3 − 2) − ĵ(−6 − 1) + k̂(4 − (−1)) = 1 î + 7 ĵ + 5 k̂.
• For coplanarity, C · (A × B) = 0:
• (3)(1) + (λ)(7) + (5)(5) = 0 ⇒ 3 + 7 λ + 25 = 0 ⇒ 7 λ = −28 ⇒ λ = −4.
Vector Addition (Doubling One Vector Doubles Resultant)
JEE Main 2026 (Jan Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q14. Two forces P and Q acting at a point have a resultant of magnitude R. If the magnitude of force Q is doubled (to 2Q), the magnitude of the new resultant is also doubled (to 2R). And if the direction of force Q is reversed (to −Q, keeping its original magnitude Q), the magnitude of the resultant again becomes 2R. What is the ratio of the magnitudes (P : Q : R)?
✔ Classic 3-Equation Vector System P : Q : R = √2 : √3 : √2 Mastered! +4 Marks
✖ Eliminate cos θ between P2 + Q2 + 2PQ cos θ = R2, P2 + 4Q2 + 4PQ cos θ = 4R2, and P2 + Q2 − 2PQ cos θ = 4R2!
Concept & Solution:• Eq (1): P2 + Q2 + 2 P Q cos θ = R2.
• Eq (2): P2 + Q2 − 2 P Q cos θ = 4 R2.
• Adding (1) and (2) eliminates cos θ: 2 P2 + 2 Q2 = 5 R2.
• Also from (3): P2 + 4 Q2 + 4 P Q cos θ = 4 R2. Adding 2×Eq(2) + Eq(3) gives 3 P2 + 6 Q2 = 12 R2 ⇒ P2 + 2 Q2 = 4 R2.
• Solving gives P2 = R2 and Q2 = (3 / 2) R2 ⇒ P : Q : R = √2 : √3 : √2.
Set 29 Grand Finale (435-PYQ Milestone: Unit Vector Perpendicular to Two Vectors)
JEE Main 2026 (Apr Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q15. What is the unit vector n̂ that is simultaneously perpendicular to both vector A = (2 î + ĵ − k̂) and vector B = (î − ĵ + 2 k̂)?
✔ 435th Question Milestone Conquered! n̂ = ± (A × B) / |A × B|! +4 Marks
✖ Compute A × B = î(2 − 1) − ĵ(4 − (−1)) + k̂(−2 − 1) = î − 5ĵ − 3k̂ (or check dot products with A and B = 0)!
Concept & Solution:• Any vector perpendicular to both A and B is parallel to their cross product A × B:
• A × B = î [ (1)(2) − (−1)(−1) ] − ĵ [ (2)(2) − (−1)(1) ] + k̂ [ (2)(−1) − (1)(1) ]
• = 1 î − 5 ĵ − 3 k̂ (Quick check: A · (A × B) = 2 − 5 + 3 = 0 ✔).
• Magnitude |A × B| = √(12 + (−5)2 + (−3)2) = √35.
• Therefore, n̂ = ± (î − 5 ĵ − 3 k̂) / √35.
Set 29 — 435-PYQ Series Completed!
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