# Projection Vector — Formula, Derivation, and Examples

TL;DR

The projection vector of \( \vec{a} \) onto \( \vec{b} \) is the "shadow" \( \vec{a} \) casts along the direction of \( \vec{b} \), given by  
\[\text{proj}_{\vec{b}}, \vec{a} = \left(\frac{\vec{a} \cdot \vec{b}}{\|\vec{b}\|^2}\right) \vec{b}\]  
This article covers the projection formula and its derivation, the difference between scalar and vector projection, what a negative projection means, and worked examples.

## What Is A Projection Vector?

The **projection vector** of \( \vec{a} \) onto \( \vec{b} \) is the component of \( \vec{a} \) that lies in the direction of \( \vec{b} \) — the vector you get by dropping a perpendicular from the tip of \( \vec{a} \) onto the line of \( \vec{b} \). It has a magnitude _and_ a direction (the direction of \( \vec{b} \)).

Two related quantities share the name "projection," and keeping them apart is half the battle:

- The **scalar projection** (also called the component) is a single signed number — _how long_ the shadow is.
- The **vector projection** is that length attached to \( \hat{b} \)'s direction — _the shadow itself_, as a vector.

You get the vector projection by multiplying the scalar projection by the unit vector \( \hat{b} \).

## What Is The Projection Vector Formula?

The **scalar projection** of \( \vec{a} \) onto \( \vec{b} \) is:

\[\text{comp}_{\vec{b}}, \vec{a} = \|\vec{a}\| \cos \theta = \frac{\vec{a} \cdot \vec{b}}{\|\vec{b}\|}\]

The **vector projection** is the scalar projection pointed along \( \hat{b} \):

\[\text{proj}_{\vec{b}}, \vec{a} = \left(\frac{\vec{a} \cdot \vec{b}}{\|\vec{b}\|^2}\right) \vec{b}\]

**Variable glossary.** \( \vec{a} \) is the vector being projected; \( \vec{b} \) is the direction projected onto; \( \theta \) is the angle between them; \( \vec{a} \cdot \vec{b} \) is the dot product; \( \|\vec{b}\| \) is the magnitude of \( \vec{b} \); \( \hat{b} = \frac{\vec{b}}{\|\vec{b}\|} \) is the unit vector along \( \vec{b} \).

Notice the two formulas differ by one factor of \( \|\vec{b}\| \) and the direction \( \vec{b} \): the scalar version divides by \( \|\vec{b}\| \) once and stops at a number; the vector version divides by \( \|\vec{b}\|^2 \) and multiplies back by \( \vec{b} \) to point the result.

## Where Does The Formula Come From?

Draw \( \vec{a} \) and \( \vec{b} \) from a common point O, with angle \( \theta \) between them. Drop a perpendicular from the tip of \( \vec{a} \) to the line carrying \( \vec{b} \), meeting it at L. The segment OLO is the length of the projection.

In right triangle OAO:

\[\cos \theta = \frac{OL}{\|\vec{a}\|} \Rightarrow OL = \|\vec{a}\| \cos \theta \]

The dot product gives a way to express \( \|\vec{a}\| \cos \theta \) without measuring the angle.  
Since \( \vec{a} \cdot \vec{b} = \|\vec{a}\| \|\vec{b}\| \cos \theta \), dividing both sides by \( \|\vec{b}\| \) isolates the scalar projection:

\[OL = \|\vec{a}\| \cos \theta = \frac{\vec{a} \cdot \vec{b}}{\|\vec{b}\|}\]

To turn that length into a vector, attach it to the unit vector \( \hat{b} \):

\[\text{proj}_{\vec{b}}, \vec{a} = \left(\frac{\vec{a} \cdot \vec{b}}{\|\vec{b}\|}\right) \hat{b} = \left(\frac{\vec{a} \cdot \vec{b}}{\|\vec{b}\|^2}\right) \vec{b}\]

## What Does A Negative Projection Mean?

The scalar projection is **signed**. When the angle \( \theta \) between the vectors is less than 90°, \( \cos \theta \) is positive and the projection points the same way as \( \vec{b} \). When \( \theta \) is greater than 90°, \( \cos \theta \) is negative, the scalar projection comes out negative, and the vector projection points _opposite_ to \( \vec{b} \). When \( \theta = 90° \) exactly, the projection is zero — the vectors are perpendicular and \( \vec{a} \) casts no shadow along \( \vec{b} \).

## Examples of Projection Vector

### Example 1

**Find the scalar projection of \( \vec{a} = 3\hat{i} + 4\hat{j} \) onto \( \vec{b} = \hat{i} \).**  
\[\vec{a} \cdot \vec{b} = (3)(1) + (4)(0) = 3,  \quad \|\vec{b}\| = 1\]
\[\text{comp}_{\vec{b}}, \vec{a} = \frac{3}{1} = 3\]  
**Final answer:** 3. Projecting onto the x-axis just reads off the x-component, as expected.

### Example 2

**Find the vector projection of \( \vec{a} = 4\hat{i} + \hat{j} \) onto \( \vec{b} = 2\hat{i} + 2\hat{j} \).**  
\[\vec{a} \cdot \vec{b} = (4)(2) + (1)(2) = 10,  \|\vec{b}\|^2 = 2^2 + 2^2 = 8\]
\[\text{proj}_{\vec{b}}, \vec{a} = \frac{10}{8}(2, \hat{i} + 2, \hat{j}) = \frac{5}{4}(2, \hat{i} + 2, \hat{j}) \]  
**Final answer:** \( \frac{5}{2}\hat{i} + \frac{5}{2}\hat{j} \).

### Example 3

**Find the scalar projection of \( \vec{a} = 4\hat{i} + 2\hat{j} + \hat{k} \) onto \( \vec{b} = 5\hat{i} - 3\hat{j} + 3\hat{k} \).**  
\[\vec{a} \cdot \vec{b} = (4)(5) + (2)(-3) + (1)(3) = 17,  \|\vec{b}\| = \sqrt{5^2 + (-3)^2 + 3^2} = \sqrt{43}\]
\[\text{comp}_{\vec{b}}, \vec{a} = \frac{17}{\sqrt{43}}\]  
**Final answer:** \( \frac{17}{\sqrt{43}} \approx 2.59 \).

### Example 4

**Find the scalar projection of \( \vec{a} = \hat{i} + 2\hat{j} \) onto \( \vec{b} = -2\hat{i} - \hat{j} \).**  
\[\vec{a} \cdot \vec{b} = (1)(-2) + (2)(-1) = -4,  \|\vec{b}\| = \sqrt{(-2)^2 + (-1)^2} = \sqrt{5}\]
\[\text{comp}_{\vec{b}}, \vec{a} = \frac{-4}{\sqrt{5}} \approx -1.79 \]  
The result is negative, so \( \vec{a} \) has a component pointing _against_ \( \vec{b} \) — the angle between them is obtuse.

### Example 5

**Project \( \vec{a} = 3\hat{i} + 4\hat{j} \) onto \( \vec{b} = -4\hat{i} + 3\hat{j} \).**  
\[\vec{a} \cdot \vec{b} = (3)(-4) + (4)(3) = 0 \]  
The dot product is zero, so the vectors are perpendicular and the projection is the zero vector.  
**Final answer:** \( \vec{0} \).

### Example 6

**A 20 N force pulls a sled along a rope at 60° above the ground. How much of the force acts horizontally (along the ground)?**  
The horizontal direction is the projection direction, and \( 
\cos 60° = \frac{1}{2} \).
\[ F_{\text{horizontal}} = \|\vec{F}\| \cos \theta = 20 \cos 60° = 10 \, \text{N} \]

## Why Projection Vectors Matter: "Aplitting A Force Into The Part That Counts"

The projection vector exists to answer one recurring question — _how much of this vector acts in that direction?_ And the answer drives a surprising amount of applied math. The deeper payoff is decomposition: every vector \( \vec{a} \) splits cleanly into the part along \( \vec{b} \) (the projection) and the part perpendicular to it.

## What Are The Most Common Mistakes With Projection Vectors?

### Mistake 1:  Using \( \|\vec{b}\| \) instead of \( \|\vec{b}\|^2 \) for the vector projection

### Mistake 2:  Projecting onto the wrong vector

### Mistake 3:  Dropping the sign of a negative projection

## Conclusion

- The **projection vector** of \( \vec{a} \) onto \( \vec{b} \) is \( \text{proj}_{\vec{b}}, \vec{a} = \left(\frac{\vec{a} \cdot \vec{b}}{\|\vec{b}\|^2}\right) \vec{b} \).
- The **scalar projection** \( \vec{a} \cdot \vec{b} \) is a signed number; the vector projection attaches that number to \( \hat{b} \).
- Projection is not symmetric: projecting \( \vec{a} \) onto \( \vec{b} \) differs from projecting \( \vec{b} \) onto \( \vec{a} \).

## A Practical Next Step

Practice these problems to solidify your understanding.