Orthogonal Projection

Core idea

Overview

The orthogonal projection of a vector v onto a vector u determines the component of v that points in the same direction as u. This process effectively maps v onto the line spanned by u, creating a new vector that is the closest point in that line to the original vector v.

When to use: Use this formula when you need to decompose a vector into parallel and perpendicular components relative to a reference vector. It is essential in the Gram-Schmidt process for building orthonormal bases and for finding the shortest distance from a point to a line.

Why it matters: Orthogonal projections are the mathematical foundation for linear regression in statistics, signal processing, and computer graphics. They allow engineers to resolve forces into specific directions and data scientists to reduce the dimensionality of complex datasets.

Symbols

Variables

c = Scalar Coefficient, u $\cdot$ v = u · v, u $\cdot$ u = u · u

c

Scalar Coefficient

Variable

u \cdot v

u · v

Variable

u \cdot u

u · u

Variable

Walkthrough

Derivation

Derivation/Understanding of Orthogonal Projection

This derivation shows how to find the component of a vector $v$ that lies along another vector $u$ , known as the orthogonal projection.

Vectors $u$ and $v$ are elements of a real inner product space (e.g., $R^{n}$ ).
The vector $u$ is non-zero, i.e., $u \neq = 0$ .

1

Define the projected vector and its properties:

We define the projection as a vector $p$ that lies along $u$ . Since it's along $u$ , it must be a scalar multiple of $u$ .

Let proj_{u} (v) = p . The vector p must be parallel to u, so p = k u for some scalar k \in R .

2

Establish the orthogonality condition:

The defining characteristic of an orthogonal projection is that the 'error' vector, $v - p$ , is perpendicular to the vector $u$ onto which $v$ is projected.

The vector from v to its projection p must be orthogonal to u . Thus, (v - p) \cdot u = 0.

3

Substitute and expand the dot product:

We replace $p$ with its expression in terms of $k$ and $u$ , then distribute the dot product to isolate the scalar $k$ .

Substitute p = k u into the orthogonality condition: (v - k u) \cdot u = 0. Using properties of the dot product: v \cdot u - k (u \cdot u) = 0.

4

Solve for the scalar k and express the projection:

By solving for $k$ , we find the scalar factor that scales $u$ to give the projection vector, thus completing the derivation.

Rearrange the equation to solve for k : k (u \cdot u) = v \cdot u ⟹ k = \frac{u \cdot v}{u \cdot u} . Substitute k back into p = k u : proj_{u} (v) = \frac{u \cdot v}{u \cdot u} u .

Result

Rearrange the equation to solve for k : k (u \cdot u) = v \cdot u ⟹ k = \frac{u \cdot v}{u \cdot u} . Substitute k back into p = k u : proj_{u} (v) = \frac{u \cdot v}{u \cdot u} u .

Source: Lay, D. C., Lay, S. R., & McDonald, J. J. (2016). Linear Algebra and Its Applications (5th ed.). Pearson.

Free formulas

Rearrangements

Solve for $c$

Orthogonal Projection

\frac{d o t U V}{d o t U U}

Start from the formula for orthogonal projection. Identify the scalar coefficient 'c' and then isolate it to express 'c' in terms of the dot products.

Difficulty: 2/5

The static page shows the finished rearrangements. The app keeps the full worked algebra walkthrough.

Visual intuition

Graph

Graph unavailable for this formula.

The graph is a linear function passing through the origin, where the scalar coefficient on the y-axis scales proportionally with the independent variable on the x-axis. This linear relationship exists because the projection formula scales the vector u by a constant ratio determined by the dot products of the input vectors.

Graph type: linear

Why it behaves this way

Intuition

Imagine vector v casting a shadow onto the line defined by vector u, where the 'light source' is perpendicular to u.

u

The reference vector defining the direction or subspace onto which another vector is projected.

This vector sets the 'target line' or 'direction' for the projection.

v

The vector being projected.

This is the vector whose component along 'u' we want to find.

u \cdot v

The dot product of vectors u and v, a scalar value representing the extent to which they point in the same direction, scaled by their magnitudes.

This quantifies the 'overlap' or 'alignment' between u and v. A positive value means they point generally in the same direction, negative means opposite, and zero means they are orthogonal.

u \cdot u

The dot product of vector u with itself, which is the squared magnitude (length) of vector u.

This term normalizes the projection, ensuring the result is scaled correctly regardless of the length of u. It effectively removes the magnitude of u from the numerator's u

\cdot

v and then reintroduces the direction the system.

\frac{u \cdot v}{u \cdot u}

A scalar coefficient that determines the 'length' and 'direction' (relative to u) of the projected vector.

This is 'how much' of v lies along u. If it's positive, the projected vector points in the same direction as u. If negative, it points opposite to u.

proj_{u} (v)

The resulting vector, which is the component of vector v that lies entirely in the direction of vector u.

This is the 'shadow' of v cast onto the line defined by u, or the part of v that is 'parallel' to u.

Signs and relationships

u · v: The dot product can be negative if the angle between vectors u and v is obtuse (greater than 90 degrees). This correctly indicates that the projection of v onto u will point in the opposite direction to u.

Free study cues

Insight

Canonical usage

All vectors involved in the projection (the vector being projected, the vector onto which it is projected, and the resulting projected vector) must share the same units.

Common confusion

A common mistake is to incorrectly assign units to the scalar factor (u · v) / (u · u), which is dimensionless, or to the final projected vector, which must retain the units of the original vectors.

Dimension note

The scalar factor (u · v) / (u · u) is dimensionless, as it is a ratio of magnitudes squared. However, the final vector proj_u(v) retains the units of the original vectors u and v.

Unit systems

$u$ Any consistent unit (e.g., meters, Newtons, or dimensionless for abstract - The unit of vector u must be consistent with vector v.

$v$ Any consistent unit (e.g., meters, Newtons, or dimensionless for abstract - The unit of vector v must be consistent with vector u.

$p r o j_{u} (v)$ Same as u and v - The projected vector retains the units of the original vectors.

One free problem

Practice Problem

In a physics simulation, a force vector v is projected onto a directional vector u. If the dot product u ⋅ v is calculated as 18 and the dot product of u with itself (u ⋅ u) is 6, what is the resulting scalar multiplier for the projection?

u · v18

u · u6

Solve for: result

Hint: Divide the dot product of the two vectors by the dot product of the reference vector u with itself.

The full worked solution stays in the interactive walkthrough.

Where it shows up

Real-World Context

Finding the component of a gravitational force acting parallel to the surface of an inclined plane.

Study smarter

Tips

Ensure the reference vector u is non-zero to avoid division by zero.
The result variable here represents the scalar coefficient that scales vector u.
Remember that u ⋅ u is the same as the squared magnitude of u.

Avoid these traps

Common Mistakes

Using the magnitude of u instead of the dot product u · u (the squared magnitude) in the denominator.
Confusing the vector being projected (v) with the vector defining the direction (u).

Keep going

Related Formulas

Common questions

Frequently Asked Questions

This derivation shows how to find the component of a vector $v$ that lies along another vector $u$, known as the orthogonal projection.

Use this formula when you need to decompose a vector into parallel and perpendicular components relative to a reference vector. It is essential in the Gram-Schmidt process for building orthonormal bases and for finding the shortest distance from a point to a line.

Orthogonal projections are the mathematical foundation for linear regression in statistics, signal processing, and computer graphics. They allow engineers to resolve forces into specific directions and data scientists to reduce the dimensionality of complex datasets.

Using the magnitude of u instead of the dot product u · u (the squared magnitude) in the denominator. Confusing the vector being projected (v) with the vector defining the direction (u).

Finding the component of a gravitational force acting parallel to the surface of an inclined plane.

Ensure the reference vector u is non-zero to avoid division by zero. The result variable here represents the scalar coefficient that scales vector u. Remember that u ⋅ u is the same as the squared magnitude of u.

References

Sources

Linear Algebra and Its Applications by David C. Lay
Introduction to Linear Algebra by Gilbert Strang
Wikipedia: Vector projection
Wikipedia: Projection (linear algebra)
Lay, David C. Linear Algebra and Its Applications. 5th ed. Pearson, 2016.
Wikipedia: Projection (linear algebra). Wikimedia Foundation. Available at: https://en.wikipedia.org/wiki/Projection_(linear_algebra)
Lay, D. C., Lay, S. R., & McDonald, J. J. (2016). Linear Algebra and Its Applications (5th ed.). Pearson.

Overview

Variables

Derivation

Define the projected vector and its properties:

Establish the orthogonality condition:

Substitute and expand the dot product:

Solve for the scalar k and express the projection:

Rearrangements

Graph

Intuition

Insight

Practice Problem

Real-World Context

Tips

Common Mistakes

Related Formulas

Gram-Schmidt Orthogonalization

Matrix Trace

Rank-Nullity Theorem

Frequently Asked Questions

Sources