Projection Vector — Formula, Derivation, and Examples
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.