➡️ Vectors — Laws of Addition, Resolution, Scalar and Vector Products

Grade XI Physics - NEB Curriculum

Unit 2: Vectors

Many physical quantities need a direction as well as a size. Vectors are the mathematical tool for handling them. This unit covers the triangle, parallelogram and polygon laws, the resolution of vectors with unit vectors, and the scalar (dot) and vector (cross) products.

Scalars and Vectors

ScalarVector
Has magnitude onlyHas magnitude and direction, and obeys the laws of vector addition
Added by ordinary algebraAdded by the triangle / parallelogram / polygon laws
Mass, time, distance, speed, work, energy, temperatureDisplacement, velocity, acceleration, force, momentum, torque

A vector is written in bold (A) or with an arrow (A⃗). Its magnitude is |A| = A. A vector is drawn as a directed line segment: the length (to scale) shows the magnitude, and the arrowhead shows the direction.

Some types of vectors:

  • Equal vectors: same magnitude and same direction.
  • Negative of a vector (−A): same magnitude, opposite direction.
  • Null (zero) vector: magnitude zero; direction undefined.
  • Unit vector: magnitude one; it shows direction only.
  • Position vector: locates a point relative to the origin.
  • Collinear vectors lie along the same line (or parallel lines); coplanar vectors lie in one plane.

⚠️ Not every quantity with a direction is a vector:

Electric current has a direction but adds as a scalar, because it does not obey the vector addition laws. Whether a quantity is a vector depends on how it combines, not on whether an arrow can be drawn for it.

2.1 Triangle, Parallelogram and Polygon Laws of Vectors

2.1(a) Triangle Law of Vector Addition

Statement: If two vectors are represented in magnitude and direction by the two sides of a triangle taken in the same order, their resultant is represented by the third side taken in the opposite order.

OABR = A + B
Figure 2.1: Triangle law. The tail of B is placed at the head of A; R = A + B joins the tail of A to the head of B

2.1(b) Parallelogram Law of Vector Addition

Statement: If two vectors acting at a point are represented in magnitude and direction by the two adjacent sides of a parallelogram drawn from that point, their resultant is represented by the diagonal of the parallelogram drawn from the same point.

θαOPQCDA (= OP)BRAD = B cos θCD = B sin θ
Figure 2.2: Parallelogram law. Vectors A and B act at O with angle θ between them; R is the diagonal OC. CD is drawn perpendicular to OP extended

Magnitude and direction of the resultant

In Fig 2.2, OP = A, OQ = B, ∠POQ = θ and the resultant is OC = R. Draw CD perpendicular to OP extended. In the right-angled triangle ADC (AC = OQ = B):

AD = B cos θ , CD = B sin θ

In the right-angled triangle ODC:

OC² = OD² + CD² = (OA + AD)² + CD²
R² = (A + B cos θ)² + (B sin θ)²
R² = A² + 2AB cos θ + B² (cos²θ + sin²θ)
R = √(A² + B² + 2AB cos θ)

If the resultant makes angle α with A:

tan α = CD / OD = B sin θ / (A + B cos θ)
Angle θ between A and BResultant RMeaning
0° (same direction)A + BMaximum
90° (perpendicular)√(A² + B²)tan α = B / A
180° (opposite)|A − B|Minimum
Equal magnitudes, A = B2A cos(θ/2)Acts along the bisector, α = θ/2

Properties of vector addition:

  • Commutative: A + B = B + A
  • Associative: (A + B) + C = A + (B + C)
  • The magnitude of the resultant lies between |A − B| and A + B.

2.1(c) Polygon Law of Vector Addition

Statement: If a number of vectors are represented in magnitude and direction by the sides of a polygon taken in the same order, their resultant is represented by the closing side of the polygon taken in the opposite order.

ABCDR = A + B + C + D (closing side)O
Figure 2.3: Polygon law. Vectors A, B, C and D are placed head to tail; R joins the starting point to the end point

The polygon law is the triangle law applied repeatedly. If the vectors form a closed polygon (the head of the last vector returns to the tail of the first), the resultant is zero.

2.1(d) Subtraction of Vectors

Subtracting B from A means adding the negative of B to A: A − B = A + (−B).

AB−BA − BOA − B = A + (−B)
Figure 2.4: Subtraction. −B has the magnitude of B but opposite direction; A − B joins the head of B to the head of A

Example 2.1

Two forces of 3 N and 4 N act at a point at right angles. Find the resultant and its direction.

R = √(3² + 4² + 2×3×4×cos 90°) = √25 = 5 N
tan α = 4 sin 90° / (3 + 4 cos 90°) = 4/3
α = 53.1° from the 3 N force

Example 2.2

Two forces of 10 N each act at 120° to each other. Find the resultant.

R = √(100 + 100 + 2×10×10×cos 120°) = √(200 − 100) = 10 N
α = θ/2 = 60° from either force

2.2 Resolution of Vectors; Unit Vectors

2.2(a) Resolution of a Vector

Splitting a vector into two or more vectors that together have the same effect is called resolution. It is the reverse of addition. The most useful split is into two rectangular components, along perpendicular axes.

θxyOAAx = A cos θAy = A sin θ
Figure 2.5: Rectangular components of A along the x and y axes
Ax = A cos θ
Ay = A sin θ
A = √(Ax² + Ay²) , tan θ = Ay / Ax

θ is the angle between the vector and the x-axis. If the angle is measured from the y-axis instead, the sine and cosine swap.

2.2(b) Unit Vectors

A unit vector has magnitude 1 and no units. It gives only direction. The unit vector along A is:

 = A / |A| so that A = |A| Â

Three special unit vectors point along the positive x, y and z axes of a right-handed system: i, j and k.

XYZijk|i| = |j| = |k| = 1
Figure 2.6: Unit vectors i, j, k along the mutually perpendicular axes X, Y, Z
A = Ax i + Ay j + Az k
|A| = √(Ax² + Ay² + Az²)

2.2(c) Addition and Subtraction Using Components

For A = Axi + Ayj + Azk and B = Bxi + Byj + Bzk, add or subtract the matching components:

A ± B = (Ax ± Bx) i + (Ay ± By) j + (Az ± Bz) k

Direction cosines:

If A makes angles α, β, γ with the x, y, z axes, then l = cos α = Ax/A, m = cos β = Ay/A, n = cos γ = Az/A, and

l² + m² + n² = 1

2.2(d) Resolving a Weight on an Inclined Plane

θmgmg sin θmg cos θθ
Figure 2.7: Weight mg of a block on a plane inclined at θ, resolved along and perpendicular to the plane
Component along the plane (down the slope) = mg sin θ
Component perpendicular to the plane = mg cos θ

The angle between the weight and the perpendicular to the plane equals the angle θ of the incline.

Example 2.3

A force of 50 N acts at 30° above the horizontal. Find its horizontal and vertical components.

Fx = 50 cos 30° = 43.3 N
Fy = 50 sin 30° = 25 N

Example 2.4

Find the magnitude and unit vector of A = 2i + 3j + 6k.

|A| = √(4 + 9 + 36) = 7
 = (2i + 3j + 6k) / 7

Example 2.5

If A = 2i + 3j and B = 4i − j, find A + B, A − B and |A + B|.

A + B = 6i + 2j
A − B = −2i + 4j
|A + B| = √(36 + 4) = 6.32

2.3 Scalar and Vector Products

2.3(a) Scalar (Dot) Product

The scalar product of two vectors A and B is a scalar equal to the product of their magnitudes and the cosine of the angle θ between them:

A · B = AB cos θ
B cos θ (component of B along A)θABA · B = A (B cos θ)= AB cos θ
Figure 2.8: A · B = A × (projection of B on A) = A (B cos θ)

Properties:

  • Commutative: A · B = B · A
  • Distributive: A · (B + C) = A · B + A · C
  • If A ⊥ B (θ = 90°), then A · B = 0. If A ∥ B (θ = 0°), A · B = AB.
  • A · A = A², and the dot product of any vector with itself is the square of its magnitude.
  • Unit vectors: i · i = j · j = k · k = 1 and i · j = j · k = k · i = 0.

Scalar product in component form

Expand (Axi + Ayj + Azk) · (Bxi + Byj + Bzk) using the unit-vector results. Only the terms with i·i, j·j and k·k survive:

A · B = AxBx + AyBy + AzBz
cos θ = (A · B) / (AB)

Physical examples: Work done W = F · d = Fd cos θ; power P = F · v.

2.3(b) Vector (Cross) Product

The vector product of A and B is a vector C:

A × B = AB sin θ n̂

where θ is the angle between A and B and n̂ is the unit vector perpendicular to the plane of A and B. Its magnitude is |A × B| = AB sin θ.

Direction: right-hand rule

  • Right-hand screw rule: rotate a right-handed screw from A towards B through the smaller angle; the direction in which the screw advances is the direction of A × B.
  • Right-hand fingers rule: curl the fingers of the right hand from A towards B; the thumb points along A × B.
θABC = A × Bshaded area = |A × B|(perpendicular to the planeof A and B, upward)
Figure 2.9: C = A × B is perpendicular to the plane of A and B. Its magnitude equals the area of the parallelogram on A and B

Properties:

  • Not commutative: A × B = −(B × A)
  • Distributive: A × (B + C) = A × B + A × C
  • If A ∥ B (θ = 0° or 180°), A × B = 0. In particular A × A = 0.
  • If A ⊥ B, |A × B| = AB (maximum).
  • |A × B| = area of the parallelogram on A and B; area of the triangle = ½ |A × B|.
ijki × j = k , j × k = i , k × i = j(clockwise = +, anticlockwise = −)
Figure 2.10: Cross products of unit vectors. Following the arrows gives a positive result; against them, negative (j × i = −k)
i × i = j × j = k × k = 0
i × j = k , j × k = i , k × i = j
j × i = −k , k × j = −i , i × k = −j

Vector product in component form (determinant)

A × B = | i j k ; Ax Ay Az ; Bx By Bz |
A × B = (AyBz − AzBy) i − (AxBz − AzBx) j + (AxBy − AyBx) k

(The first line is the 3×3 determinant with rows i j k; Ax Ay Az; Bx By Bz, expanded along the first row.)

Physical examples: Torque τ = r × F (magnitude rF sin θ); angular momentum L = r × p; magnetic force on a moving charge F = q (v × B).

Scalar productVector product
ResultScalarVector
MagnitudeAB cos θAB sin θ
Commutative?YesNo (A × B = −B × A)
Zero whenθ = 90° (perpendicular)θ = 0° or 180° (parallel)
Maximum whenθ = 0°θ = 90°
ExampleWork = F · dTorque = r × F

Example 2.6

Show that A = 2i + 3j + k and B = i − 2j + 4k are perpendicular.

A · B = (2)(1) + (3)(−2) + (1)(4) = 2 − 6 + 4 = 0

The dot product is zero, so the vectors are perpendicular.

Example 2.7

Find the angle between A = i + j and B = j + k.

A · B = 0 + 1 + 0 = 1 ; |A| = |B| = √2
cos θ = 1 / (√2 × √2) = 1/2 ⇒ θ = 60°

Example 2.8

A force F = (5i + 3j) N moves a body through d = (4i + 2j) m. Find the work done.

W = F · d = (5)(4) + (3)(2) = 26 J

Example 2.9

Find A × B for A = 2i + 3j + k and B = i − j + 2k.

A × B = (3×2 − 1×(−1)) i − (2×2 − 1×1) j + (2×(−1) − 3×1) k
= 7i − 3j − 5k

Example 2.10

A force of 10 N acts at the end of a 2 m spanner at 30° to the spanner. Find the torque.

τ = rF sin θ = 2 × 10 × sin 30° = 10 N m

Key Formulas and Summary

Addition of Vectors:

R = √(A² + B² + 2AB cos θ)
tan α = B sin θ / (A + B cos θ) (α from A)
Rmax = A + B (θ = 0°) , Rmin = |A − B| (θ = 180°)

Components and Unit Vectors:

Ax = A cos θ , Ay = A sin θ
A = Axi + Ayj + Azk , |A| = √(Ax² + Ay² + Az²)
 = A / |A|

Products:

A · B = AB cos θ = AxBx + AyBy + AzBz
|A × B| = AB sin θ
i · i = 1, i · j = 0 ; i × j = k, j × k = i, k × i = j

Practice Numerical Problems

Attempt each problem before reading the answer.

1. Two forces of 6 N and 8 N act at a point at 90°. Find the resultant and its direction from the 6 N force.

Answer: R = √(36 + 64) = 10 N; tan α = 8/6, α = 53.1°.

2. Two vectors of equal magnitude A act at 60°. Find the magnitude and direction of the resultant.

Answer: R = √(A² + A² + 2A² cos 60°) = √3 A; α = θ/2 = 30° from either vector.

3. Find the resultant of 5 N and 12 N forces when the angle between them is (a) 0°, (b) 180°, (c) 90°.

Answer: (a) 17 N; (b) 7 N; (c) √(25 + 144) = 13 N.

4. Resolve a 100 N force acting at 37° above the horizontal into horizontal and vertical components (sin 37° = 0.6, cos 37° = 0.8).

Answer: Fx = 100 × 0.8 = 80 N; Fy = 100 × 0.6 = 60 N.

5. A 10 kg block rests on a smooth plane inclined at 30°. Find the components of its weight along and perpendicular to the plane (g = 9.8 m s⁻²).

Answer: mg = 98 N. Along the plane: 98 sin 30° = 49 N; perpendicular: 98 cos 30° ≈ 84.9 N.

6. Find the unit vector along A = 3i − 4j and verify that its magnitude is 1.

Answer: |A| = 5; Â = (3i − 4j)/5 = 0.6i − 0.8j; |Â| = √(0.36 + 0.64) = 1.

7. Find the angle between A = 2i − 3j + k and B = i + j − k.

Answer: A · B = 2 − 3 − 1 = −2; |A| = √14, |B| = √3; cos θ = −2/√42 ≈ −0.309, θ ≈ 108°.

8. Find A × B and its magnitude for A = i + 2j + 3k and B = 3i − j + 2k.

Answer: A × B = (4 + 3) i − (2 − 9) j + (−1 − 6) k = 7i + 7j − 7k; magnitude = 7√3 ≈ 12.1.

9. For what value of λ are A = 2i + λj + k and B = 4i − 2j − 2k perpendicular?

Answer: A · B = 8 − 2λ − 2 = 0 ⇒ λ = 3.

10. A force of 10 N acts at 60° to the displacement of 5 m of a body. Find the work done.

Answer: W = Fd cos θ = 10 × 5 × 0.5 = 25 J.

11. Find the area of the parallelogram whose adjacent sides are A = 3i and B = 4j, and the area of the triangle on the same sides.

Answer: A × B = 12k, so parallelogram area = 12 units; triangle area = 6 units.

Conceptual and Long Answer Questions

1. Can the resultant of two vectors be smaller in magnitude than either of the vectors? Explain.
Yes. For example, 5 N and 4 N acting in opposite directions give 1 N. The resultant lies between |A − B| and A + B.
2. Is electric current a vector? Why?
No. It has direction but does not obey the vector addition laws; currents add algebraically at a junction.
3. Why is work a scalar product while torque is a vector product?
Work has no direction (energy transferred), so F · d is a scalar. Torque has a direction (the axis of rotation), so r × F is a vector.
4. When is A · B = 0 and when is A × B = 0, with both A and B non-zero?
A · B = 0 when A ⊥ B; A × B = 0 when A ∥ B (θ = 0° or 180°).
5. A vector has non-zero magnitude. Can one of its rectangular components be larger than the vector itself?
No. Ax = A cos θ and Ay = A sin θ never exceed A in magnitude.
6. State the parallelogram law and derive the magnitude and direction of the resultant. (Section 2.1)
7. Define the scalar and vector products with two examples each, and prove A × B = −B × A using the right-hand rule. (Section 2.3)

Multiple Choice Questions

1. The resultant of two vectors is maximum when the angle between them is:
(a) 0° ✓
(b) 90°
(c) 120°
(d) 180°
2. Which of the following is a vector quantity?
(a) Work
(b) Power
(c) Torque ✓
(d) Energy
3. i · j is equal to:
(a) 1
(b) 0 ✓
(c) k
(d) −1
4. j × k is equal to:
(a) −i
(b) 0
(c) i ✓
(d) 1
5. The magnitude of the vector 3i + 4j is:
(a) 7
(b) 5 ✓
(c) 1
(d) 25
6. The component of a vector A making angle θ with the x-axis along the y-axis is:
(a) A cos θ
(b) A tan θ
(c) A sin θ ✓
(d) A / sin θ
7. A · (A × B) equals:
(a) A²B
(b) AB
(c) 1
(d) 0 ✓
8. Two forces of 3 N and 4 N cannot give a resultant of:
(a) 1 N
(b) 5 N
(c) 7 N
(d) 9 N ✓

Grade XI Physics - Vectors | NEB Curriculum