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How to Find the Angle Between Two Vectors?

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  • Last Updated : 01 Feb, 2022

Physical quantities that have both directions and magnitude are vectors. A vector with magnitude equal to one and direction equal to one is a unit vector “û.”. That is the lower-case alphabet with a “hat” circumflex. In this way, vectors are described by arrows, have initial points and terminal points, and were developed over a period of 200 years. Vectors can be used to represent physical quantities, such as displacement, velocity, acceleration, etc. 

Angle Between Two Vectors

Vectors with angle θ between them

A vector’s angle between its tails is equal to its angle between two vectors. It can be obtained using a dot product (scalar product) or cross product (vector product). Note that the angle between the two vectors remains between 0° and 180°. The angle between vectors can be found by using two methods. But the most commonly used formula for finding an angle between two vectors involves the scalar product. 

Finding angle using  Scalar (dot) product 

Two vectors combined into a scalar product give you a number. Scalar products can be used to define the relationships between energy and work. In mathematics, a scalar product is used to represent the work done by a force (which is a vector) in dispersing (which is a vector) an object. The scalar product is represented by a dot (.). Let,

Dot product be (a.b)

Magnitude of vector a = |a|

Magnitude of vector b = |b|

Angle between the vectors is θ = Cos-1 [(a · b) / (|a| |b|)]

When two vectors are connected by a dot product, the direction of the angle ፀ does not matter. The angle ፀ can be measured by the difference between either vector since Cos ፀ = Cos (-ፀ) = Cos (2π – ፀ).

Finding angle using cross (vector) product

A cross product may also be known as a vector product. It is a form of vector multiplication that takes place between two vectors that have different kinds or natures. When two vectors are multiplied with each other and the resulting product is also a vector quantity, the resulting vector is called the cross product of two vectors or the vector product. Multiplication of two vectors yields vector products with a direction perpendicular to each vector. Let,

Cross product be (a × b)

Magnitude of vector a = |a|

Magnitude of vector b = |b|.

|a × b| = |a| |b| sin θ

Angle between the vectors is θ = Sin-1 [|a × b|  / (|a| |b|)]

Sample Problems

Question 1: Find the angle between two vectors a = {4, 5} and b = {5, 4}.

Solution:

  1. Finding dot product ( a.b) = 4 × 5 + 5 × 4 = 40.
  2. Finding vectors magnitude, |a| = \sqrt{((4×4)+(5×5))} = √41, |b| = \sqrt{((5×5)+(4×4))} = √41.
  3. Angle between vectors, θ = Cos-1 [(a · b) / (|a| |b|)] , θ = Cos-1 [(40) / (√41 × √41)]

Angle between a and b, 

θ = Cos-1 [(40) / (41)]

Question 2: Find the angle between two vectors a = {2, 2} and b = {1, 1}.

Solution:

  1. Finding dot product (a.b) = 2 × 1 + 2 × 1 = 4.
  2. Finding vectors magnitude, |a| = \sqrt{((2×2)+(2×2))} = √8, |b| = \sqrt{((1×1)+(1×1))} = √2.
  3. Angle between vectors, θ = Cos-1 [(a · b) / (|a| |b|)], θ = Cos-1 [(4) / (√8 × √2)].

Angle between a and b,

θ = Cos-1 [(4) / (4)] 

θ = Cos-1 [1] = 0°.

This means both vectors are overlapping each other and are in same direction.

Question 3: Find the angle between two vectors a = i + 2j – k and b = 2i + 4j – 2k.

Solution:

  1. Finding dot product (a.b) = 1 × 2 + 2 × 4 + (-1) × (-2) = 2 + 8 + 2 = 12.
  2. Finding vectors magnitude, |a| = \sqrt{((1×1)+(2×2)+(-1×-1))} = √6, |b| = \sqrt{((2×2)+(4×4)+(-2×-2))} = √24.
  3. Angle between vectors, θ = Cos-1 [(a · b) / (|a| |b|)] , θ = Cos-1 [(12) / (√6 × √24)].

Angle between a and b,

θ = Cos-1 [(12) / (12)]

θ = Cos-1 [1] = 0°.

This means both vectors are overlapping each other and are in same direction.

Question 4: Find the angle between two vectors a = i + 2j – k and b = 4j – 2k.

Solution:

Vector b can be written as, b = 0i + 4j – 2k.

  1. Finding dot product (a.b) = 1 × 0 + 2 × 4 + (-1) × (-2) = 0 + 8 + 2 = 10.
  2. Finding vectors magnitude, |a| = \sqrt{((1×1)+(2×2)+(-1×-1))} = √6, |b| = \sqrt{((0×0)+(4×4)+(-2×-2))} = √20.
  3. Angle between vectors, θ = Cos-1 [(a · b) / (|a| |b|)] , θ = Cos-1 [(10) / (√6 × √20)].

Angle between a and b,

θ = Cos-1 [(10) / (√120)].

Question 5: Find the angle between two vectors a = {1, -3} and b = {-3, 1}.

Solution:

  1. Finding cross product magnitude |a × b| = √(0)² + (0)² + (-8)² = 8.
  2. Finding vectors magnitude, | a| = \sqrt{((1×1)+(-3×-3))} = √10, |b| = \sqrt{((-3×-3)+(1×1))} = √10.
  3. Angle between vectors, θ = Sin-1 [(|a × b|) / (|a| |b|)], θ = Sin-1 [(8) / (√10 × √10)]

Angle between a and b,

θ = Sin-1 [(8) / (10)] 

Question 6: Find the angle between two vectors a = -3i + j and b = -3i + j.

Solution:

  1. Finding dot product (a.b) = (-3) × (-3) + 1 × 1 = 10.
  2. Finding vectors magnitude, |a| = \sqrt{(((-3)×(-3))+(1×1))} = √10, |b| = \sqrt{(((-3)×(-3))+(1×1))} = √10
  3. Angle between vectors, θ = Cos-1 [(a · b) / (|a| |b|)], θ = Cos-1 [(10) / (√10 × √10)].

Angle between a and b,

θ = Cos-1 [(10) / (10)]

θ = Cos-1 [1] = 0°

As both vectors are having same value and direction so they are same vectors hence having angle 0 between them.


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