Projecting vectors onto a line

Geometric meaning of projection in $2$ dimensions

Explanation

The scalar projection of a vector $\vec{v}$ onto another vector $\vec{w}$ tells us how much of $\vec{v}$ points in the direction of $\vec{w}$ , measured as a signed length.

Scalar projection

Definition

The scalar projection is defined as:

\text{proj}_{\vec{w}} \vec{v} = \|\vec{v}\| \cos(\theta)

where $\theta$ is the angle between $\vec{v}$ and $\vec{w}$ . The line defined by $\vec{w}$ must pass through the origin, since projection formulas assume both vectors are positioned with their tails at the origin.

Geometrically, to find the scalar projection, we drop a perpendicular from the tip of $\vec{v}$ to the line defined by $\vec{w}$ (which extends infinitely in both directions). The point where this perpendicular meets the line is the "shadow" of $\vec{v}$ , and the distance from the origin to this point - measured along the line - is the scalar projection.

We denote this projection as $\text{proj}_{\vec{w}} \vec{v}$ .

Have a play with the interactive visualisation below and see how the scalar projection changes as $\vec{v}$ is moved around.

0,0

A

\vec{v}

B

\vec{w}

C

{proj}_{\vec{w}} \vec{v}

θ

| \vec{v} | = 2.83

{proj}_{\vec{w}} \vec{v} = 2.00

In the example above, $\vec{w}$ points in the direction of the $x$ -axis, but in general, $\vec{v}$ and $\vec{w}$ can point in any direction. If $\vec{v}$ and $\vec{w}$ point in roughly opposite directions, the scalar projection will be negative.

To see this, suppose $\|\vec{v}\| = 3$ and $\vec{w}$ makes an angle of $120^\circ$ with $\vec{v}$ . Then we have:

\begin{align*} \text{proj}_{\vec{w}} \vec{v} &= \|\vec{v}\| \cos(120^\circ) \\ &= 3 \cos(120^\circ) \\ &= -1.5. \end{align*}

Have a play with the interactive visualisation below where both $\vec{v}$ and $\vec{w}$ are now draggable.

Notice how we drop a perpendicular from the tip of $\vec{v}$ to the line defined by $\vec{w}$ . The right angle is formed where this perpendicular meets the line, which may be "behind" the origin relative to $\vec{w}$ if the angle is greater than $90^\circ$ . This is why the scalar projection can be negative.

0,0

A

\vec{v}

\vec{w}

B

{proj}_{\vec{w}} \vec{v}

θ

| \vec{v} | = 3.00

{proj}_{\vec{w}} \vec{v} = - 2.57

Intuition

Tip

A negative scalar projection means that $\vec{v}$ points at least partially in the opposite direction to $\vec{w}$ . The projection only considers movement along $\vec{w}$ , so if $\vec{v}$ "moves backward" along $\vec{w}$ , the result is negative.

Key points

Key Point

The scalar projection measures the component of $\vec{v}$ along $\vec{w}$ .
It is a scalar value.
The value can be positive or negative depending on the direction.

Example

Let $\vec{v}$ be a vector with $|\vec{v}| = 9$ .

Let $\vec{w}$ be a vector that makes an angle of $\dfrac{2\pi}{3}$ with $\vec{v}$ .

Calculate $\text{proj}_{\vec{w}} \vec{v}$ .

Solution

To calculate the projection of $\vec{v}$ onto $\vec{w}$ , we use the formula:

\begin{align*} \text{proj}_{\vec{w}} \vec{v} &= |\vec{v}| \cos(\theta) \\ &= 9 \cos\left(\dfrac{2\pi}{3}\right) \\ &= 9 \left(-\dfrac{1}{2}\right) \\ &= -\dfrac{9}{2}. \end{align*}

Practice questions

3 questions

Let $\vec{v}$ be a vector with $|\vec{v}| = 6$ .

Let $\vec{w}$ be a vector that makes an angle of $\dfrac{\pi}{4}$ with $\vec{v}$ .

Calculate $\text{proj}_{\vec{w}} \vec{v}$ .

Select the correct answer:

$2 \sqrt{2}$

$-\dfrac{6}{\sqrt{2}}$

$\dfrac{6}{\sqrt{2}}$

$-2 \sqrt{2}$

+ 2 more questions

Scalar and vector projection in $2$ dimensions

Explanation

Now that we’ve understood the scalar projection as the signed length of the shadow of $\vec{v}$ on the direction of $\vec{w}$ , let’s connect this idea to the dot product. This will also set us up to properly describe what that blue arrow in our visualisations has actually been representing all along.

The scalar projection of $\vec{v}$ onto $\vec{w}$ is the signed length of the "shadow" of $\vec{v}$ in the direction of $\vec{w}$ .

Also recall the dot product formula:

\vec{v} \cdot \vec{w} = \|\vec{v}\| \|\vec{w}\| \cos(\theta)

Dividing both sides by $\|\vec{w}\|$ isolates $\|\vec{v}\| \cos(\theta)$ , which is the scalar projection:

\|\vec{v}\| \cos(\theta) = \dfrac{\vec{v} \cdot \vec{w}}{\|\vec{w}\|}

Written in the projection notation, this is the formula we use from here on.

Scalar projection from the dot product

Theorem

Let $\vec{v}$ and $\vec{w}$ be non-zero vectors. The scalar projection of $\vec{v}$ onto $\vec{w}$ is

\text{proj}_{\vec{w}} \vec{v} = \dfrac{\vec{v} \cdot \vec{w}}{\|\vec{w}\|}

This formula is especially useful because it works in any dimension and doesn’t require us to find the angle. The dot product measures how much two vectors "align", and dividing by $\|\vec{w}\|$ ensures the result is independent of the length of $\vec{w}$ .

So far we’ve only measured the length of the shadow. But in our diagrams above we also drew the blue arrow lying along $\vec{w}$ , and strictly speaking that arrow isn’t the scalar projection at all.

Vector projection

Definition

Let $\vec{v}$ and $\vec{w}$ be non-zero vectors. The vector projection of $\vec{v}$ onto $\vec{w}$ , written $\text{proj}_{\vec{w}}^{\text{vec}} \vec{v}$ , is the vector lying on the line defined by $\vec{w}$ whose signed length is the scalar projection of $\vec{v}$ onto $\vec{w}$ .

So the scalar projection tells us how long the shadow is, a number, while the vector projection is the blue arrow itself.

The vector projection formula

Theorem

The vector projection is given by:

\text{proj}_{\vec{w}}^{\text{vec}} \vec{v} = \left( \dfrac{\vec{v} \cdot \vec{w}}{\|\vec{w}\|^2} \right) \vec{w}.

In words: the vector projection is simply the scalar projection multiplied by the unit vector in the direction of $\vec{w}$ .

Derivation

Details

Recall that the scalar projection gives us the signed length of the shadow:

\text{proj}_{\vec{w}} \vec{v} = \dfrac{\vec{v} \cdot \vec{w}}{\|\vec{w}\|}

To turn this length into a vector, we need to multiply by a unit vector in the direction of $\vec{w}$ :

\text{unit vector in direction of } \vec{w} = \dfrac{\vec{w}}{\|\vec{w}\|}

So, the vector projection is:

\text{proj}_{\vec{w}}^{\text{vec}} \vec{v} = \left( \dfrac{\vec{v} \cdot \vec{w}}{\|\vec{w}\|} \right) \dfrac{\vec{w}}{\|\vec{w}\|}

This simplifies to:

\text{proj}_{\vec{w}}^{\text{vec}} \vec{v} = \left( \dfrac{\vec{v} \cdot \vec{w}}{\|\vec{w}\|^2} \right) \vec{w}

Both quantities describe the same shadow, so their sizes must agree.

Norm of the vector projection

Theorem

For non-zero vectors $\vec{v}$ and $\vec{w}$ , the norm (length) of the vector projection is always the absolute value of the scalar projection:

\left\| \text{proj}_{\vec{w}}^{\text{vec}} \vec{v} \right\| = \left| \text{proj}_{\vec{w}} \vec{v} \right|

Since the scalar projection measures the signed length of the shadow of $\vec{v}$ along $\vec{w}$ , and the vector projection points in the correct direction, taking the norm of the vector projection simply gives the (unsigned) length of that shadow.

This is a useful way to check our calculations - if we take the norm of the vector projection, it should match the absolute value of the scalar projection.

Key points

Key Point

The scalar projection gives the signed length of the shadow of $\vec{v}$ in the direction of $\vec{w}$ .
The vector projection gives the actual vector along $\vec{w}$ .
Using the dot product, both can be computed directly from the dot product, without needing to find the angle.

Example

Let $\vec{v} = \begin{bmatrix} 2 \\ 3 \end{bmatrix}$ and $\vec{w} = \begin{bmatrix} 4 \\ 0 \end{bmatrix}$ .

Calculate both the scalar projection and the vector projection of $\vec{v}$ onto $\vec{w}$ .

Solution

First, we compute the dot product:

\begin{align*} \vec{v} \cdot \vec{w} &= 2 \times 4 + 3 \times 0 \\ &= 8 + 0 \\ &= 8. \end{align*}

Next, we find the norm of $\vec{w}$ :

\begin{align*} \|\vec{w}\| &= \sqrt{4^2 + 0^2} \\ &= \sqrt{16} \\ &= 4. \end{align*}

The scalar projection is then:

\begin{align*} \text{proj}_{\vec{w}} \vec{v} &= \dfrac{\vec{v} \cdot \vec{w}}{\|\vec{w}\|} \\ &= \dfrac{8}{4} \\ &= 2. \end{align*}

For the vector projection, we use the formula (noting that $\|\vec{w}\|^2 = 16$ ):

\begin{align*} \text{proj}_{\vec{w}}^{\text{vec}} \vec{v} &= \left( \dfrac{\vec{v} \cdot \vec{w}}{\|\vec{w}\|^2} \right) \vec{w} \\ &= \left( \dfrac{8}{16} \right) \begin{bmatrix} 4 \\ 0 \end{bmatrix} \\ &= \left( \dfrac{1}{2} \right) \begin{bmatrix} 4 \\ 0 \end{bmatrix} \\ &= \begin{bmatrix} 2 \\ 0 \end{bmatrix}. \end{align*}

Practice questions

4 questions

Suppose $\vec{a}$ and $\vec{b}$ are non-zero vectors in $\mathbb{R}^2$ .

Consider the following statements about the scalar and vector projection of $\vec{a}$ onto $\vec{b}$ :

A. The scalar projection of $\vec{a}$ onto $\vec{b}$ can be negative, zero, or positive depending on the angle between the vectors.

B. The vector projection of $\vec{a}$ onto $\vec{b}$ is always a scalar multiple of $\vec{b}$ .

C. If the scalar projection is zero, the vector projection is the zero vector.

D. The norm (length) of the vector projection is always equal to the scalar projection.

E. The vector projection of $\vec{a}$ onto $\vec{b}$ always points in the same direction as $\vec{b}$ .

Which of the above statements are true?

Select the correct answer:

Statements A and C

Statements B, C, D and E

Statements B and C

Statements A, B and C

+ 3 more questions

Generalisation to $N-$ dimensional vectors

Explanation

So far, we’ve focused on projections in two dimensions, but the same ideas and formulas apply in any number of dimensions. The dot product and norm work exactly the same way for $N$ -dimensional vectors, so the projection formulas generalise directly.

Projection formulas in

n

dimensions

Theorem

Let $\vec{v}$ and $\vec{w}$ be vectors in $\mathbb{R}^n$ with $\vec{w}$ non-zero. The scalar projection of $\vec{v}$ onto $\vec{w}$ is

\text{proj}_{\vec{w}} \vec{v} = \dfrac{\vec{v} \cdot \vec{w}}{\|\vec{w}\|}

and the vector projection of $\vec{v}$ in the direction of $\vec{w}$ is

\text{proj}_{\vec{w}}^{\text{vec}} \vec{v} = \left( \dfrac{\vec{v} \cdot \vec{w}}{\|\vec{w}\|^2} \right) \vec{w}

These formulas work for any non-zero direction vector $\vec{w}$ in any dimension - we do not need to normalise $\vec{w}$ first, as the formulas automatically account for its length.

Example

Let $\vec{v} = \begin{bmatrix} -1 \\ 4 \\ 5 \end{bmatrix}$ and $\vec{w} = \begin{bmatrix} 0 \\ 2 \\ -2 \end{bmatrix}$ .

Calculate both the scalar projection and the vector projection of $\vec{v}$ onto $\vec{w}$ .

Solution

First, we compute the dot product:

\begin{align*} \vec{v} \cdot \vec{w} &= (-1)(0) + (4)(2) + (5)(-2) \\ &= 0 + 8 - 10 \\ &= -2. \end{align*}

Next, we find the norm of $\vec{w}$ :

\begin{align*} \|\vec{w}\| &= \sqrt{0^2 + 2^2 + (-2)^2} \\ &= \sqrt{0 + 4 + 4} \\ &= \sqrt{8} \\ &= 2\sqrt{2}. \end{align*}

The scalar projection is then:

\begin{align*} \text{proj}_{\vec{w}} \vec{v} &= \dfrac{\vec{v} \cdot \vec{w}}{\|\vec{w}\|} \\ &= \dfrac{-2}{2\sqrt{2}} \\ &= -\dfrac{1}{\sqrt{2}}. \end{align*}

For the vector projection, we use the formula (noting that $\|\vec{w}\|^2 = (2\sqrt{2})^2 = 8$ ):

\begin{align*} \text{proj}_{\vec{w}}^{\text{vec}} \vec{v} &= \left( \dfrac{\vec{v} \cdot \vec{w}}{\|\vec{w}\|^2} \right) \vec{w} \\ &= \left( \dfrac{-2}{8} \right) \begin{bmatrix} 0 \\ 2 \\ -2 \end{bmatrix} \\ &= \left( -\dfrac{1}{4} \right) \begin{bmatrix} 0 \\ 2 \\ -2 \end{bmatrix} \\ &= \begin{bmatrix} 0 \\[10pt] -\dfrac{1}{2} \\[10pt] \dfrac{1}{2} \end{bmatrix}. \end{align*}

Practice questions

3 questions

Let $\vec{v} = \begin{bmatrix} 3 \\ -6 \\ 4 \\ 2 \end{bmatrix}$ and $\vec{w} = \begin{bmatrix} 2 \\ 0 \\ -1 \\ 5 \end{bmatrix}$ .

What is the scalar projection of $\vec{v}$ onto $\vec{w}$ ?

Select the correct answer:

$\dfrac{21}{\sqrt{30}}$

$\dfrac{8}{\sqrt{30}}$

$\dfrac{12}{\sqrt{30}}$

$\dfrac{7}{\sqrt{30}}$

+ 2 more questions

Master the linear algebra that powers modern ML

Projecting vectors onto a line

Explanation

Example

Explanation

Example

Explanation

Example

Vectors in Euclidean space