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Vector

NEET > Physics > Physical World and Measurement > Fundamental Mathematics and Vector > Vector

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NEET Physics — Fundamental Mathematics and Vector

Vector – Complete Notes, Revision, Important Questions & Downloads

Vector analysis in NEET Physics spans ten core subtopics: Definition and types of vectors (scalars vs vectors, unit/zero/polar/axial vectors), Triangle Law of Vector Addition (magnitude R = √(A²+B²+2AB cos θ) and direction formula), Parallelogram Law of Vector Addition (diagonal-based resultant with special cases at 0°, 90°, 180°), Polygon Law of Vector Addition (n-sided polygon closure), Subtraction of Vectors (A−B as A+(−B) with modified magnitude formula), Resolution of Vector Into Rectangular Components (R_x = R cos θ, R_y = R sin θ), Rectangular Components of 3-D Vector (direction cosines l²+m²+n²=1), Scalar Product of Two Vectors (A·B = AB cos θ, used in work and power), Vector Product of Two Vectors (A×B = AB sin θ n̂, used in torque and angular momentum), and Lami's Theorem (equilibrium condition A/sin α = B/sin β = C/sin γ). NEET tests this topic through resultant-magnitude calculations, component-resolution numericals, and dot/cross product applications in force and torque problems — for instance, finding the angle between two vectors given A·B = AB/2 immediately yields θ = 60°.

⬇ Download Notes PDFView Important Questions →
23 SubtopicsCore Mechanics ToolNCERT Class 11
Expected QuestionsQ
1–2
NEET asks 1–2 direct vector questions per year on resultant magnitude, component resolution, or dot/cross product applications. Vectors also appear indirectly in 30–40% of mechanics numericals.
Time Required⏱
6–8 hours
Two sessions for vector addition laws and subtraction, two sessions for resolution and 3-D components, two sessions for scalar and vector products plus Lami's theorem, followed by mixed-problem practice.
Difficulty⚡
Medium
The addition laws and resolution are straightforward; difficulty rises with cross product direction determination and applying Lami's theorem to three-force equilibrium.
NRI USA Curriculum GapUS
Medium–High
US AP Physics 1 introduces vectors qualitatively but does not require formal triangle/parallelogram law derivations, direction cosines, or determinant-based cross product computation. NRI students need explicit drill on these algebraic techniques.
23Subtopics
15+Practice Questions
4Free Downloads
6–8 hrsPrep Time
⬇ Get Free Downloads

NEET Weightage — Vector

Fundamental Mathematics and Vector (Chapter 0)
NEET YearQuestions from this TopicBarMarks
20241
 
1 Q
4
20232
 
2 Q
8
20221
 
1 Q
4
20211
 
1 Q
4
20201
 
1 Q
4
20191
 
1 Q
4
6-Year Total (2019–2024)7–9 28–36
Resultant magnitude R = √(A²+B²+2AB cos θ) is tested as a direct numerical — NEET gives two force magnitudes and the angle, asks for the resultant. Memorise the special cases: θ=0° gives A+B, θ=90° gives √(A²+B²), θ=180° gives |A−B|.
Dot product A·B = AB cos θ is frequently embedded in work-energy problems — NEET may give F and s in component form and ask W = F·s = F_x s_x + F_y s_y.

Cross product direction via the right-hand rule appears in torque (τ = r×F) and magnetic force (F = qv×B) — NEET tests whether students can determine the correct perpendicular direction.
📊
~1.2
Avg Questions / Year
🎯
28–36
Total Marks (6 yrs)
📈
Mixed
Pattern
⚠️
Medium
Difficulty

Exam Strategy for Vector in NEET Physics

1

Memorise the resultant formula and its three special cases Commit R = √(A²+B²+2AB cos θ) and tan α = B sin θ/(A+B cos θ) to memory. Before solving, check if θ is 0°, 90°, or 180° — these shortcuts halve your calculation time. The trap: forgetting that at θ=180°, R = |A−B|, not A−B (magnitude is always positive).

2

Drill component resolution in both 2-D and 3-D For any vector R at angle θ to the x-axis: R_x = R cos θ, R_y = R sin θ. In 3-D, verify l²+m²+n²=1 as a sanity check. The trap: swapping cos and sin when the angle is measured from the y-axis instead of the x-axis.

3

Distinguish dot product from cross product by output type Dot product yields a scalar (work, power, projection); cross product yields a vector (torque, angular momentum, magnetic force). When NEET asks 'find the work done', use F·s. When it asks 'find the torque', use r×F. The trap: computing A×B when the problem asks for the component of A along B (which requires A·B/B).

4

Apply Lami's theorem only when exactly three concurrent coplanar forces are in equilibrium Check the three conditions: (1) exactly three forces, (2) concurrent (meeting at one point), (3) in equilibrium (net force = 0). Then A/sin α = B/sin β = C/sin γ where α, β, γ are the angles opposite to A, B, C respectively. The trap: using the angle between the vectors instead of the angle opposite to each vector.

Download Study Notes — Vector

PDF · Cheat Sheet · MCQ Set · PYQ
📘
Vector — Full Notes
Complete notes covering all 10 subtopics: vector types, triangle/parallelogram/polygon addition laws, vector subtraction, 2-D and 3-D resolution, direction cosines, scalar product, vector product with determinant method, and Lami's theorem with worked NEET examples.
23 subtopicsWorked examplesAll formulas
Download PDF
📗
Vector — Formula Sheet
One-page formula reference: resultant magnitude and direction, component resolution equations, dot product (A·B = A_xB_x+A_yB_y+A_zB_z), cross product determinant, direction cosines relation, and Lami's theorem — key formulas, conditions, and one worked example per subtopic.
1 pageAll key formulas
Download PDF
📙
Vector — MCQ Practice
20 NEET-style MCQs testing resultant calculations, component resolution, dot product for work and angle determination, cross product for torque direction, and Lami's theorem equilibrium problems.
20 MCQsDetailed solutions
Download PDF
📕
Vector — NEET-Style PYQ Practice
Collection of NEET-style practice questions on vector addition at various angles, force resolution along inclined planes, work done using dot product in component form, and three-force equilibrium using Lami's theorem.
NEET-styleAnswer key included
Download PDF

Subtopics in Vector

2-Column Table
Column AColumn B
Definition and types of vectors↗
Triangle Law of Vector Addition↗
Parallelogram Law of Vector Addition↗
Polygon Law of Vector Addition↗
Subtraction of Vectors↗
Resolution of Vector Into Rectangular Components↗
Rectangular Components of 3-D Vector↗
Scalar Product of Two Vectors↗
Vector Product of Two Vectors↗
Lami's Theorem↗
Orthogonal unit vectors↗
Magnitude of resultant vector↗
In terms of components↗
Unit vector gives the direction of vector↗
One vector↗
Collinear vectors↗
Polar Vectors↗
Axial Vectors↗
Coplanar vector↗
Negative Vectors↗
The vector product↗
Division of vectors↗
Distance covered↗

Rapid Revision — Vector

Concept → Trap → Example

1) Definition and types of vectors

Classification Fundamentals

A vector has magnitude, direction, and obeys laws of vector algebra. Unit vector â = A/|A| has magnitude 1. Zero vector has zero magnitude and arbitrary direction.

  • Current has magnitude and direction but is a scalar because it does not obey the parallelogram law of vector addition — a standard NEET assertion-reason trap.
  • Polar vectors (displacement, force) have a point of application; axial vectors (torque, angular momentum) lie along the rotation axis per the right-hand rule.
  • NEET trap: confusing collinear vectors (angle 0° or 180° between them) with coplanar vectors (lying in the same plane) — all collinear vectors are coplanar, but not vice versa.
Example (NEET-style)If A = 3î + 4ĵ, then |A| = √(9+16) = 5 and the unit vector â = (3î + 4ĵ)/5 = 0.6î + 0.8ĵ.

2) Triangle Law of Vector Addition

Resultant Formula

R = √(A² + B² + 2AB cos θ), direction: tan α = B sin θ/(A + B cos θ), where θ is the angle between vectors A and B.

  • The two vectors must be placed tail-to-head (same order) for the triangle law to apply — the resultant is the closing side drawn from the tail of the first to the head of the second.
  • When θ = 90°, the formula reduces to the Pythagorean form R = √(A²+B²), which is the fastest shortcut in NEET.
  • NEET trap: using the wrong angle — θ in the formula is the angle between the vectors when placed tail-to-tail, not the interior angle of the triangle.
Example (NEET-style)Two forces of 3 N and 4 N act at 60° to each other. R = √(9+16+2×3×4×cos 60°) = √(25+12) = √37 ≈ 6.08 N.

3) Parallelogram Law of Vector Addition

Adjacent-Side Resultant

Two vectors as adjacent sides of a parallelogram → resultant is the diagonal. Same magnitude formula: R = √(A² + B² + 2AB cos θ). Special cases: θ=0° → R=A+B; θ=180° → R=|A−B|; θ=90° → R=√(A²+B²).

  • The parallelogram law and triangle law give identical magnitude and direction — the geometric construction differs but the algebra is the same.
  • Maximum resultant occurs when vectors are parallel (θ=0°); minimum when anti-parallel (θ=180°).
  • NEET trap: forgetting that R_min = |A−B|, not A−B, and that R lies between |A−B| and A+B for any angle θ.
Example (NEET-style)Forces 5 N and 5 N at θ=120°: R = √(25+25+2×25×cos 120°) = √(50−25) = √25 = 5 N. Equal forces at 120° give R equal to either force.

4) Polygon Law of Vector Addition

Multi-Vector Summation

If (n−1) vectors form (n−1) sides of an n-sided polygon in order, the resultant is the nth (closing) side in the opposite order. R = A + B + C + D + …

  • The polygon law is a generalisation of the triangle law applied repeatedly — each successive vector is placed head-to-tail.
  • If n vectors form a closed polygon (all n sides in order), their resultant is zero — this is the equilibrium condition for concurrent forces.
  • NEET trap: concluding that three non-coplanar vectors can sum to zero — they cannot, because a polygon of three sides must be planar.
Example (NEET-style)Five coplanar forces forming a closed pentagon: A + B + C + D + E = 0. Remove E: the resultant of A + B + C + D = −E (equal magnitude, opposite direction to E).

5) Subtraction of Vectors

Reverse-Addition Method

A − B = A + (−B). Magnitude: |A − B| = √(A² + B² − 2AB cos θ). Direction: tan α₂ = B sin θ/(A − B cos θ).

  • Vector subtraction reverses B and then adds — the magnitude formula replaces +2AB cos θ with −2AB cos θ.
  • If |A+B| = |A−B|, then cos θ = 0, so θ = 90° — vectors are perpendicular. NEET uses this as a quick conceptual MCQ.
  • NEET trap: assuming |A−B| = A−B. This holds only when A and B are parallel (θ=0°); otherwise the magnitude formula must be used.
Example (NEET-style)A = 7 N, B = 3 N, θ = 60°: |A−B| = √(49+9−2×7×3×cos 60°) = √(58−21) = √37 ≈ 6.08 N.

6) Resolution of Vector Into Rectangular Components

2-D Decomposition

R = R_x î + R_y ĵ where R_x = R cos θ, R_y = R sin θ. Inverse: R = √(R_x²+R_y²), θ = tan⁻¹(R_y/R_x).

  • The angle θ is always measured from the axis along which the cosine component is taken — if measured from the y-axis, R_x = R sin θ and R_y = R cos θ.
  • On an inclined plane, resolve gravity along and perpendicular to the plane: mg sin α (along) and mg cos α (perpendicular) — this is the most-tested NEET application.
  • NEET trap: swapping sin and cos when the reference axis changes from x to y or from horizontal to the incline.
Example (NEET-style)A 10 N force at 30° to the horizontal: F_x = 10 cos 30° = 10×(√3/2) = 5√3 ≈ 8.66 N, F_y = 10 sin 30° = 5 N.

7) Rectangular Components of 3-D Vector

Direction Cosines

R = R_x î + R_y ĵ + R_z k̂. Direction cosines: l = cos α = R_x/R, m = cos β = R_y/R, n = cos γ = R_z/R. Constraint: l² + m² + n² = 1.

  • The constraint cos²α + cos²β + cos²γ = 1 serves as a quick verification — if your direction cosines do not satisfy it, recheck the component calculation.
  • Position vector of point (x,y,z) is r = xî + yĵ + zk̂; displacement vector from (x₁,y₁,z₁) to (x₂,y₂,z₂) is Δr = (x₂−x₁)î + (y₂−y₁)ĵ + (z₂−z₁)k̂.
  • NEET trap: computing the magnitude as R_x + R_y + R_z instead of √(R_x²+R_y²+R_z²) — vector magnitudes require the Pythagorean sum, not arithmetic addition.
Example (NEET-style)R = 2î + 3ĵ + 6k̂: |R| = √(4+9+36) = √49 = 7. Direction cosines: l = 2/7, m = 3/7, n = 6/7. Check: 4/49+9/49+36/49 = 49/49 = 1.

8) Scalar Product of Two Vectors

Dot Product & Applications

A·B = AB cos θ = A_xB_x + A_yB_y + A_zB_z. Properties: commutative (A·B = B·A), î·î = 1, î·ĵ = 0. Applications: W = F·s, P = F·v.

  • If A·B = 0 and neither A nor B is zero, then θ = 90° — the vectors are perpendicular. This is the orthogonality test NEET frequently uses.
  • The component form A_xB_x + A_yB_y + A_zB_z is faster than AB cos θ when vectors are given in î, ĵ, k̂ notation.
  • NEET trap: computing work as |F|×|s| without cos θ — work uses the dot product, so W = Fs cos θ, not Fs.
Example (NEET-style)F = 3î + 4ĵ and s = 2î + 5ĵ. Work W = F·s = 3×2 + 4×5 = 6 + 20 = 26 J.

9) Vector Product of Two Vectors

Cross Product & Direction

A×B = AB sin θ n̂. Determinant: A×B = |î ĵ k̂; A_x A_y A_z; B_x B_y B_z|. Non-commutative: A×B = −B×A. Self product: A×A = 0.

  • The direction of A×B is given by the right-hand screw rule: curl fingers from A to B through the smaller angle, and the thumb points along A×B.
  • |A×B| gives the area of the parallelogram formed by A and B; ½|A×B| gives the area of the triangle.
  • NEET trap: ignoring the non-commutative property — A×B = −(B×A), so reversing the order flips the direction. This matters in torque (r×F ≠ F×r) and magnetic force (qv×B ≠ qB×v).
Example (NEET-style)A = 2î + 3ĵ, B = î + 2ĵ. A×B = (2×2−3×1)k̂ = (4−3)k̂ = k̂. Area of parallelogram = |A×B| = 1 sq unit.

10) Lami's Theorem

Three-Force Equilibrium

If A + B + C = 0 (equilibrium), then A/sin α = B/sin β = C/sin γ, where α, β, γ are angles opposite to vectors A, B, C respectively.

  • Lami's theorem applies only to exactly three concurrent coplanar forces in equilibrium — verify all three conditions before applying.
  • The angles α, β, γ are the angles between the other two forces (opposite angles), not the angles each force makes with a reference axis.
  • NEET trap: confusing the angle between two forces with the angle opposite to a force — if forces B and C make angle α between them, then α is the angle opposite to force A in the Lami's construction.
Example (NEET-style)A 10 kg lamp hangs from two strings making 30° and 60° with the ceiling. Weight W = 100 N. By Lami's theorem: T₁/sin 150° = T₂/sin 120° = 100/sin 90°. So T₁ = 100 sin 150° = 50 N, T₂ = 100 sin 120° = 50√3 ≈ 86.6 N.

US Curriculum Gaps — Vector for NEET Physics

NRI students from US high schools may find these specific gaps when preparing for NEET Physics vectors.

Formal Vector Addition Laws and Derivations (not required in AP Physics 1)

US AP Physics 1 introduces vectors through graphical tip-to-tail addition and basic component methods, but does not require students to derive the magnitude formula R = √(A²+B²+2AB cos θ) from the triangle or parallelogram law. NEET expects both the derivation and rapid application of this formula with its direction formula tan α = B sin θ/(A+B cos θ).

  • AP Physics 1 uses graphical vector addition and calculators for components — NEET requires algebraic derivation by hand.
  • Direction cosines (l, m, n) and the constraint l²+m²+n² = 1 are not covered in any standard US high school physics course.
  • Practice deriving R and α from the triangle law diagram before applying the formula numerically.

Cross Product Determinant Method and Right-Hand Screw Rule (limited in AP Physics C: Mechanics)

AP Physics C introduces the cross product conceptually for torque and angular momentum, but typically relies on |A×B| = AB sin θ without requiring the full 3×3 determinant expansion. NEET problems frequently present vectors in component form (î, ĵ, k̂) and expect the determinant method for computing the cross product, plus the right-hand screw rule for direction.

  • US courses compute torque magnitude as rF sin θ but rarely require the î(A_yB_z−A_zB_y) − ĵ(A_xB_z−A_zB_x) + k̂(A_xB_y−A_yB_x) expansion.
  • Lami's theorem for three-force equilibrium is not part of any standard US physics curriculum — NRI students encounter it for the first time in NEET prep.
  • Drill the determinant cross-product on 5–10 examples until the cofactor expansion is automatic.

NEET-Style Practice Questions — Vector

8 NEET-style practice questions
1Two forces of magnitude 5 N and 12 N act on a particle. If the angle between them is 90°, the magnitude of their resultant is:NEET-style practice
7 N
13 N
17 N
8.5 N
Using the resultant formula R = √(A²+B²+2AB cos θ) with A=5 N, B=12 N, θ=90°: R = √(25+144+2×5×12×cos 90°) = √(169+0) = √169 = 13 N. At θ=90°, the cross term vanishes, giving the Pythagorean result. Option (a) 7 N = |A−B| is the minimum resultant (at θ=180°). Option (c) 17 N = A+B is the maximum resultant (at θ=0°). Option (d) 8.5 N is the arithmetic mean, which has no physical basis. The key step is recognising that cos 90° = 0 simplifies the formula to R = √(A²+B²).
2If A = 3î + 4ĵ and B = −4î + 3ĵ, then A·B equals:NEET-style practice
24
−24
0
25
A·B = A_xB_x + A_yB_y = (3)(−4) + (4)(3) = −12 + 12 = 0. Since A·B = 0 and neither vector is zero, the vectors are perpendicular (θ = 90°). Verification: |A| = 5, |B| = 5, A·B = 25 cos θ = 0 gives cos θ = 0, confirming θ = 90°. Option (a) incorrectly sums |A_xB_x| + |A_yB_y| = 12+12 = 24. Option (b) incorrectly takes −(|A_xB_x| + |A_yB_y|). Option (d) computes |A|×|B| = 25, which is the product of magnitudes, not the dot product.
3The cross product of A = 2î + 3ĵ + k̂ and B = î − ĵ + 2k̂ is:NEET-style practice
7î − 3ĵ − 5k̂
7î + 3ĵ + 5k̂
−7î + 3ĵ + 5k̂
7î − 3ĵ + 5k̂
Using the determinant method: 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) = 7î − 3ĵ − 5k̂. Step by step: î-component = A_yB_z − A_zB_y = 6−(−1) = 7; ĵ-component = −(A_xB_z − A_zB_x) = −(4−1) = −3; k̂-component = A_xB_y − A_yB_x = −2−3 = −5. Option (b) ignores the negative signs in the ĵ and k̂ cofactors. Option (c) reverses the î sign. Option (d) gets only the k̂ sign wrong.
4A force F = 6î + 2ĵ − 3k̂ N acts on a particle that undergoes displacement s = 2î − 3ĵ + k̂ m. The work done by the force is:NEET-style practice
3 J
−3 J
9 J
0 J
Work W = F·s = F_xs_x + F_ys_y + F_zs_z = (6)(2) + (2)(−3) + (−3)(1) = 12 − 6 − 3 = 3 J. The dot product accounts for all three components: the positive x-contribution (12 J) is partially offset by the negative y-contribution (−6 J) and negative z-contribution (−3 J). Option (b) incorrectly reverses the overall sign. Option (c) sums absolute values, ignoring the sign convention of the dot product. Option (d) assumes perpendicular F and s, which is not the case here. This problem tests the component-form dot product formula — a frequent NEET numerical pattern for work-energy problems.
5If |A + B| = |A − B|, then the angle between A and B is:NEET-style practice
0°
60°
90°
180°
Squaring both sides: A²+B²+2AB cos θ = A²+B²−2AB cos θ. Simplifying: 4AB cos θ = 0. Since A ≠ 0 and B ≠ 0, cos θ = 0, giving θ = 90°. This is a classic NEET conceptual MCQ that tests the algebraic relationship between vector addition and subtraction. Option (a) gives |A+B| = A+B and |A−B| = |A−B|, which are equal only if B=0. Option (b) gives cos 60° = 0.5 ≠ 0. Option (d) gives |A+B| = |A−B| only if A=0. The key insight is that equality of sum and difference magnitudes implies perpendicularity.
6A vector R = 4î + 3ĵ makes an angle θ with the x-axis. The value of cos θ is:NEET-style practice
3/5
4/5
3/4
4/3
|R| = √(16+9) = √25 = 5. The direction cosine with respect to the x-axis: cos θ = R_x/|R| = 4/5. This applies the definition of direction cosines to a 2-D case. Option (a) gives sin θ = R_y/|R| = 3/5, confusing the sine and cosine. Option (c) gives R_y/R_x = 3/4, which is tan θ, not cos θ. Option (d) gives R_x/R_y = 4/3, which is cot θ. The distinction between cos θ, sin θ, and tan θ in terms of components is a standard NEET vector trap.
7Three concurrent coplanar forces 3 N, 4 N, and 5 N are in equilibrium. The angle between the 3 N and 4 N forces is:NEET-style practice
90°
120°
150°
60°
For equilibrium, the resultant of the 3 N and 4 N forces must equal 5 N (to balance the third force). Using R² = A²+B²+2AB cos θ: 25 = 9+16+24 cos θ → 24 cos θ = 0 → cos θ = 0 → θ = 90°. This confirms the 3-4-5 Pythagorean triplet relationship: 3²+4² = 5². The three forces form a right triangle when placed head-to-tail, with the right angle between the 3 N and 4 N forces. Option (b) 120° would give R = √(9+16−12) = √13 ≈ 3.6 N ≠ 5. Option (c) 150° would give R = √(25−24×(√3/2)) ≈ 1.6 N ≠ 5. Option (d) 60° would give R = √(25+12) = √37 ≈ 6.1 N ≠ 5.
8The magnitude of the resultant of two equal forces F each, when the angle between them is 120°, is:NEET-style practice
F
2F
F√3
F/2
R = √(F²+F²+2F² cos 120°) = √(2F²+2F²×(−1/2)) = √(2F²−F²) = √(F²) = F. Two equal forces at 120° produce a resultant equal in magnitude to either force. Option (b) 2F occurs only when θ=0° (parallel forces). Option (c) F√3 occurs when θ=60°: R = √(2F²+2F²×0.5) = √(3F²) = F√3. Option (d) F/2 has no standard geometric basis. This is a frequently asked NEET pattern — commit the result 'equal forces at 120° give R = F' to memory.

Practice Problems — Vector in Physics

Click "Reveal Answer" after attempting
1Two vectors A and B have magnitudes 6 and 10 respectively. If A×B has magnitude 60, find the dot product A·B.
0
30
60
30√3
👁 Reveal Answer
Option (a): 0. |A×B| = AB sin θ = 60 → 60 sin θ = 60 → sin θ = 1 → θ = 90°. Therefore A·B = AB cos θ = 60 cos 90° = 0. When the cross product magnitude equals the product of individual magnitudes, the vectors are perpendicular, making the dot product zero.
2A particle moves from position r₁ = 2î + 3ĵ − k̂ to r₂ = 5î − ĵ + 2k̂ under a constant force F = 4î + 2ĵ + k̂ N. Calculate the work done.
7 J
3 J
−5 J
12 J
👁 Reveal Answer
Option (a): 7 J. Displacement s = r₂ − r₁ = (5−2)î + (−1−3)ĵ + (2−(−1))k̂ = 3î − 4ĵ + 3k̂. Work W = F·s = (4)(3) + (2)(−4) + (1)(3) = 12 − 8 + 3 = 7 J. The positive x-contribution (12 J) dominates, partially offset by the negative y-contribution (−8 J), and augmented by the small z-contribution (3 J). Option (b) incorrectly drops one component. Option (c) would result from using F_z = −3 instead of +1. Option (d) uses only the x-component.
3Find the area of the triangle formed by vectors A = î + 2ĵ + 3k̂ and B = 3î − ĵ + 2k̂.
7√3 / 2 sq units
7√3 sq units
7 sq units
49/2 sq units
👁 Reveal Answer
Option (a): 7√3/2 sq units. A×B = î(2×2 − 3×(−1)) − ĵ(1×2 − 3×3) + k̂(1×(−1) − 2×3) = î(4+3) − ĵ(2−9) + k̂(−1−6) = 7î + 7ĵ − 7k̂. |A×B| = √(49+49+49) = √147 = 7√3. Area of triangle = ½|A×B| = 7√3/2 sq units. The cross product gives a vector perpendicular to both A and B, and its magnitude equals the area of the parallelogram spanned by A and B; half that is the triangle area.
4The unit vector perpendicular to both A = 2î + ĵ − k̂ and B = î − ĵ + k̂ is:
(2ĵ + k̂)/√5
(−3ĵ − 3k̂)/(3√2)
Cannot be determined
(2k̂ − ĵ)/√5
👁 Reveal Answer
Option (b) in simplified form. A×B = î(1×1−(−1)(−1)) − ĵ(2×1−(−1)×1) + k̂(2×(−1)−1×1) = î(1−1) − ĵ(2+1) + k̂(−2−1) = 0î − 3ĵ − 3k̂ = −3ĵ − 3k̂. |A×B| = √(0+9+9) = 3√2. Unit vector n̂ = (A×B)/|A×B| = (−3ĵ−3k̂)/(3√2) = (−ĵ−k̂)/√2. This is the vector perpendicular to both A and B. The formula n̂ = (A×B)/|A×B| always gives a unit vector perpendicular to the plane of A and B.
5If two vectors A = 3î − 2ĵ + k̂ and B = 2î + ĵ − 4k̂ are the adjacent sides of a parallelogram, find the area of the parallelogram.
7√6 sq units
7√3 sq units
√294 / 2 sq units
14 sq units
👁 Reveal Answer
Option (a): 7√6 sq units. A×B = î((−2)(−4)−(1)(1)) − ĵ((3)(−4)−(1)(2)) + k̂((3)(1)−(−2)(2)) = î(8−1) − ĵ(−12−2) + k̂(3+4) = 7î + 14ĵ + 7k̂. |A×B| = √(49+196+49) = √294 = 7√6. Area of parallelogram = |A×B| = 7√6 ≈ 17.15 sq units. Option (b) confuses √3 with √6. Option (c) is ½|A×B|, which is the triangle area, not the parallelogram area. Option (d) sums magnitudes of cross product components (7+14+7)/2 = 14, which is algebraically meaningless.

Physics — Vector Revision Checklist

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Frequently Asked Questions — Vector in NEET Physics

Notes · Downloads · Revision · Important Questions
Why is electric current a scalar even though it has a direction?
A quantity must satisfy three conditions to be a vector: it must have magnitude, direction, AND obey the laws of vector algebra (specifically the parallelogram law of addition). Electric current has magnitude and direction (from positive to negative terminal), but two currents meeting at a junction follow Kirchhoff's current law (algebraic sum = 0), not the parallelogram law. When two 3 A currents meet at a junction, the outgoing current is 6 A regardless of the angle between the wires — this is scalar addition, not vector addition.
What is the difference between the triangle law and the parallelogram law of vector addition?
Both laws give exactly the same resultant R = √(A²+B²+2AB cos θ) with the same direction formula. The difference is geometric construction only: the triangle law places vectors tail-to-head and draws the closing side, while the parallelogram law places vectors tail-to-tail as adjacent sides and draws the diagonal. The parallelogram law is more convenient when forces act from the same point, which is the typical NEET scenario.
How do I decide whether to use dot product or cross product in a NEET problem?
Check the expected output type: if the answer is a scalar (work, power, energy, flux, projection), use the dot product A·B = AB cos θ. If the answer is a vector with a specific direction (torque, angular momentum, magnetic force, area vector), use the cross product A×B = AB sin θ n̂. The key mnemonic: 'scalar answer = dot, vector answer = cross.' In component form: dot gives a number (A_xB_x+A_yB_y+A_zB_z), cross gives a vector (determinant expansion).
Why is A×B ≠ B×A, and when does this matter in NEET?
The cross product is anti-commutative: A×B = −(B×A). The magnitudes are identical, but the directions are exactly opposite. This matters in torque (τ = r×F, not F×r), angular momentum (L = r×p), and magnetic force (F = qv×B, not qB×v). In NEET, reversing the order gives the wrong direction for the resulting vector, which changes the physical answer (e.g., torque clockwise vs counterclockwise).
When can three vectors have a zero resultant?
Three vectors can have a zero resultant only if they are coplanar (lie in the same plane) and can form a closed triangle when placed head-to-tail. Specifically: if A + B + C = 0, then C = −(A + B), meaning the third vector is the negative of the resultant of the other two. Three non-coplanar vectors can never sum to zero because no vector in 3-D space can simultaneously close a triangle that extends into three dimensions. Two unequal vectors also cannot sum to zero.
How do I quickly find the angle between two vectors given in component form?
Compute cos θ = (A·B)/(|A||B|) = (A_xB_x+A_yB_y+A_zB_z)/(√(A_x²+A_y²+A_z²) × √(B_x²+B_y²+B_z²)). If the result is 0, θ=90°. If the result is 1, θ=0° (parallel). If the result is −1, θ=180° (anti-parallel). For NEET, the most common answers are θ = 0°, 60°, 90°, 120°, or 180° — check if cos θ matches 1, 1/2, 0, −1/2, or −1.
What is the physical significance of the magnitude of the cross product?
|A×B| = AB sin θ gives the area of the parallelogram formed with A and B as adjacent sides. Half of this (½|A×B|) gives the area of the triangle. In physics: torque magnitude |τ| = |r×F| = rF sin θ represents the rotational effect of force F at distance r when the angle between r and F is θ. The sin θ factor means the rotational effect is maximum when the force is perpendicular to the position vector.
What are direction cosines and how are they tested in NEET?
Direction cosines l, m, n are the cosines of the angles α, β, γ that a vector makes with the x, y, and z axes respectively: l = R_x/R, m = R_y/R, n = R_z/R. They always satisfy l²+m²+n² = 1. NEET may ask: 'A vector makes angles 60°, 45° with x and y axes. Find the angle with the z-axis.' Solution: cos²60° + cos²45° + cos²γ = 1 → 1/4 + 1/2 + cos²γ = 1 → cos²γ = 1/4 → γ = 60°.
How does Lami's theorem relate to the triangle law of vector addition?
Lami's theorem is a direct consequence of the triangle law applied to three forces in equilibrium. If A + B + C = 0, the three vectors form a closed triangle. By the sine rule in this triangle: A/sin α = B/sin β = C/sin γ, where α, β, γ are the angles of the triangle (each angle is the supplement of the angle between the corresponding pair of forces). Lami's theorem is faster than resolving into components when exactly three concurrent forces are in equilibrium and their angles are known.
Can the resultant of two vectors be smaller than both individual vectors?
Yes, when the angle θ between the vectors is obtuse (90° < θ ≤ 180°). For example, two vectors of magnitude 5 N and 4 N at θ=150°: R = √(25+16+2×20×cos 150°) = √(41−20√3) ≈ √(41−34.64) ≈ √6.36 ≈ 2.52 N, which is less than both 5 N and 4 N. The minimum possible resultant is |A−B| = 1 N (at θ=180°). The resultant equals either vector when θ=120° and the vectors have equal magnitude.
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Definition and types of vectors

Triangle Law of Vector Addition

Parallelogram Law of Vector Addition

Polygon Law of Vector Addition

Subtraction of Vectors

Resolution of Vector Into Rectangular Components

Rectangular Components of 3-D Vector

Scalar Product of Two Vectors

Vector Product of Two Vectors

Lami's Theorem

Orthogonal unit vectors

Magnitude of resultant vector

In terms of components

Unit vector gives the direction of vector

One vector

Collinear vectors

Polar Vectors

Axial Vectors

Coplanar vector

Negative Vectors

The vector product

Division of vectors

Distance covered

Subtopics

Definition and types of vectors

Triangle Law of Vector Addition

Parallelogram Law of Vector Addition

Polygon Law of Vector Addition

Subtraction of Vectors

Resolution of Vector Into Rectangular Components

Rectangular Components of 3-D Vector

Scalar Product of Two Vectors

Vector Product of Two Vectors

Lami's Theorem

Orthogonal unit vectors

Magnitude of resultant vector

In terms of components

Unit vector gives the direction of vector

One vector

Collinear vectors

Polar Vectors

Axial Vectors

Coplanar vector

Negative Vectors

The vector product

Division of vectors

Distance covered

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Definition and types of vectors

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