Vector Projection (original) (raw)

Last Updated : 4 Dec, 2025

Vector projection is a fundamental concept in physics and mathematics that describes how one vector influences another along a specific direction. It can be visualised as the shadow that one vector casts onto another when light is shone perpendicular to the second vector.

When working with two vectors, \overrightarrow{a} and\overrightarrow{b}, the projection of \overrightarrow{a} onto \overrightarrow{b} tells us how much of vector \overrightarrow{a} lies in the direction of \overrightarrow{b}.

Vector projection actually comes in two forms:

**1. Scalar Projection

This is the length of the projection — a number that tells us how much of \overrightarrow{a} points along \overrightarrow{b}. It is given by:

\text{Scalar projection of } \ \overrightarrow{a} \text{ on } \overrightarrow{b} = |\overrightarrow{a}| \cos \theta

where θ is the angle between vectors a and b.

**2. Vector Projection

This is the actual vector that represents the projection of \overrightarrow{a} onto \overrightarrow{b}. It gives both the magnitude and the direction (same as \overrightarrow{b}). It is given by:

\text{Vector projection of } \ \overrightarrow{a} \text{ on } \overrightarrow{b} = (\frac{\overrightarrow{a} \cdot \overrightarrow{b}}{|\overrightarrow{b}|^2 })\overrightarrow{b}

**Example of Vector Projection

**Pushing a Box on the Floor: Imagine you’re pushing a box at an angle to the ground using a force F.

Not all of your applied force moves the box forward; only the horizontal component of the force (the projection of F on the horizontal direction) contributes to motion. The vertical part just presses the box into the floor. This horizontal projection is Fcos⁡θ

Vector Projection Formula

If \vec A is represented as A and \vec B is represented as B, The Vector Projection of A on B is given as the product of A with Cos θ where θ is the angle between A and B. The Projection Vector obtained so is a scalar multiple of A and has a direction in the direction of B.

**Projection of Vector a on Vector b Proj\ _ba= \frac {a.b}{|b|}

a.b = Dot product of \vec a and \vec b

|b| = magnitude of \vec b

Derivation of Vector Projection Formula

Let us assume, OP = \vec A and OQ = \vec B and the angle between OP and OQ is θ. Drawn PN perpendicular to OQ.

In the right triangle OPN, Cos θ = ON/OP

⇒ ON = OP Cos θ
⇒ ON = |\vec A| Cos θ

ON is the projection vector of \vec A on \vec B

\vec A.\vec B = |\vec A||\vec B|cos \theta
⇒ \vec A.\vec B = |\vec B(|\vec A||cos \theta)
⇒ \vec A.\vec B = |\vec B|ON
⇒ ON = \frac{\vec A.\vec B}{|\vec B|}
Hence, the ON = |\vec A|.\hat B

Thus the Vector Projection of \vec A on \vec B is given as \frac{\vec A.\vec B}{|\vec B|}

The Vector Projection of \vec B on \vec A is given as \frac{\vec A.\vec B}{|\vec A|}

Vector Projection Important Terms

To find the vector projection we need to learn to find the angle between two vectors and also to calculate the dot product between two vectors.

Angle Between Two Vectors

The angle between the two vectors is given as the inverse of the cosine of the dot product of two vectors divided by the product of the magnitude of two vectors.

Let's say we have two vectors \vec A and \vec B angle between them is θ

⇒ cos θ = \frac{\vec A.\vec B}{|A|.|B|}
⇒ θ = cos-1\frac{\vec A.\vec B}{|A|.|B|}

Dot Product of Two Vectors

Let's say we have two vectors \vec A and \vec Bdefined as \vec A = a_1\hat i + a_2\hat j + a_3\hat k and \vec B = b_1\hat i + b_2\hat j + b_3\hat k then the dot product between them is given as

\vec A.\vec B = (a_1\hat i + a_2\hat j + a_3\hat k)(b_1\hat i + b_2\hat j + b_3\hat k)
⇒ \vec A.\vec B= a1b1 + a2b2 +a3b3

Related Articles:

• Vector Addition
• Unit Vector
• Vector Algebra
• Linear Algebra

Applications and Significance of Vector Projection

• **Work and Force Analysis
  Vector projection helps in finding the component of a force that acts in a particular direction. For example, when pushing an object at an angle, only the projected part of the force contributes to its movement along the surface.
• **Computer Graphics and Visualization
  In 3D graphics, projection is used to simulate lighting, shadows, and reflections. By projecting light vectors onto surfaces, realistic rendering effects are achieved.
• **Game and Animation Design
  Motion simulation in games and animations relies on projecting movement vectors onto terrain surfaces, ensuring natural and realistic character movement.
• **Navigation and Path Optimization
  GPS systems use vector projection to compute the shortest and most accurate path between two locations by projecting displacement vectors onto the Earth’s surface.

Vector Projection in Real-World

• **GPS Navigation
  Used to project displacement vectors between locations onto the Earth’s surface to determine optimal travel routes.
• **Sports Analytics
  Helps in analyzing player and ball movements by projecting motion vectors onto the field or court to study speed, direction, and performance.
• **Renewable Energy (Wind Turbines)
  Engineers project wind velocity vectors onto turbine blade planes to find the most effective orientation for maximum energy generation.
• **Augmented Reality (AR) Applications
  Vector projection is used to align virtual elements accurately with physical environments, making digital overlays appear realistic.

Vector Projection Formula Examples

**Example 1. Find the projection of a vector4\hat i + 2\hat j + \hat k on 5\hat i -3\hat j + 3\hat k.

**Solution:

Here, \vec{a}=4\hat i + 2\hat j + \hat k \\\vec{b}=5\hat i -3\hat j + 3\hat k .

We know, projection of Vector a on Vector b = \frac{\vec{a}.\vec{b}}{|b|}

\dfrac{(4.(5) + 2(-3) + 1.(3))}{|\sqrt{5^2 + (-3)^2 + 3^2}|}\\=\dfrac{17}{\sqrt{43}}

**Example 2. Find the projection of the vector 5\hat i + 4\hat j + \hat k on 3\hat i + 5\hat j - 2\hat k

**Solution:

Here, \vec{a}=5\hat i + 4\hat j + \hat k \\\vec{b}=3\hat i + 5\hat j - 2\hat k.

We know, projection of Vector a on Vector b = \frac{\vec{a}.\vec{b}}{|b|}

\dfrac{(5.(3) + 4(5) + 1.(-2))}{|\sqrt{3^2 + 5^2 + (-2)^2}|}\\=\dfrac{33}{\sqrt{38}}

**Example 3. Find the projection of the vector 5\hat i - 4\hat j + \hat k on 3\hat i - 2\hat j + 4\hat k

**Solution:

Here, \vec{a}=5\hat i - 4\hat j + \hat k \\\vec{b}=3\hat i - 2\hat j + 4\hat k.

We know, projection of Vector a on Vector b = \frac{\vec{a}.\vec{b}}{|b|}

\dfrac{(5.(3) + ((-4).(-2)) + 1.(4))}{|\sqrt{3^2 + (-2)^2 + (4)^2}|}\\=\dfrac{27}{\sqrt{29}}

**Example 4. Find the projection of the vector 2\hat i - 6\hat j + \hat kon8\hat i - 2\hat j + 4\hat k.

**Solution:

Here, \vec{a}=2\hat i - 6\hat j + \hat k \\\vec{b}=8\hat i - 2\hat j + 4\hat k

We know, projection of Vector a on Vector b = \frac{\vec{a}.\vec{b}}{|b|}

\dfrac{(2.(8) + ((-6).(-2)) + 1.(4))}{|\sqrt{8^2 + (-2)^2 + (4)^2}|}\\=\dfrac{32}{\sqrt{84}}

**Example 5. Find the projection of the vector 2\hat i - \hat j + 5\hat kon4\hat i - \hat j + \hat k.

**Solution:

Here, \vec{a}=2\hat i - \hat j + 5\hat k \\\vec{b}=4\hat i - \hat j + \hat k.

We know, projection of Vector a on Vector b = \frac{\vec{a}.\vec{b}}{|b|}

\dfrac{(2.(4) + ((-1).(-1)) + 5.(1))}{|\sqrt{4^2 + (-1)^2 + (1)^2}|}\\=\dfrac{14}{\sqrt{18}}

**Example 6: Find the angle between the vectors \vec{a}=4\hat i + 3\hat j - \hat k \\\vec{b}=2\hat i - \hat j + 2\hat k

**Solution :

\vec{a}=4\hat i + 3\hat j - \hat k \\\vec{b}=2\hat i - \hat j + 2\hat k

cos θ = \frac{\vec A.\vec B}{|A|.|B|}

a.b = (4 x 2) + (3 x (-1)) + (( -1) x 2)
= 8 - 3 - 2 = 3

|a| = √ (4) 2 + (3) 2 + (- 1) 2
|a| = √ 16 + 9 + 1 = √ 26

|b| = √ (2) 2 + (- 1) 2 + (2) 2
|b| = √ 4 + 1 + 4 =|b|
= √ 9 = 3

Now computig value cos θ

cos θ = a.b/|a||b|
= 3 /3√26
=1/√26

θ = cos-1 1/√26
θ =78.7

Unsolved Examples on Vector Projection

**Question 1: Find the projection of vector \overrightarrow{a} = 6\hat{i}-2\hat{j}+3\hat{k} on the vector \overrightarrow{b} = 3\hat{i}+4\hat{j}-\hat{k}.

**Question 2: If the project of \overrightarrow{A} on \overrightarrow{B} is 5 units, and |\overrightarrow{A}| = 10, find the angle between \overrightarrow{A} and \overrightarrow{B.}

**Question 3: The projection of vector \overrightarrow{a} = 2\hat{i}-\hat{j}+2\hat{k} on vector \overrightarrow{b} = \hat{i}+2\hat{j}+2\hat{k} is equal to the projection of \overrightarrow{b} on \overrightarrow{a}.

Find the ration of magnitudes |\overrightarrow{a}|: |\overrightarrow{b}|.

**Question 4: Two vectors are given as \overrightarrow{p} = 3\hat{i}+2\hat{j}+6\hat{k} and \overrightarrow{q} = x\hat{i}-3\hat{j}+2\hat{k}.

If the projection of \overrightarrow{p} on \overrightarrow{q} is zero, find the value of x.

**Question 5: Find the projection of vector \overrightarrow{a} = 7\hat{i}-5\hat{j}+2\hat{k} on a unit vector in the direction of \overrightarrow{b} = 3\hat{i}+2\hat{j}+6\hat{k}.

**Example 6: Let \overrightarrow{a} = \hat{i}+2\hat{j}+2\hat{k} and \overrightarrow{b} = 2\hat{i}+\hat{j}+2\hat{k}.

Find the projection vector (not just scalar) of \overrightarrow{a} on \overrightarrow{b}, and state its direction.