Vector Operations in Space: Learn It 3

Projections

Vector addition combines vectors to create a resultant vector. But what if you need to work in reverse—breaking a vector into its component parts? Vector projections solve this problem by showing how much of one vector acts in the direction of another.

Consider a child pulling a wagon handle at a 55° angle. Only part of that pulling force actually moves the wagon forward—the horizontal component. Projections help us find exactly how much force is working in the direction we care about.

This figure is the image of a wagon with a handle. The handle is represented by the vector “F.” The angle between F and the horizontal direction of the wagon is 55 degrees.
Figure 1. When a child pulls a wagon, only the horizontal component of the force propels the wagon forward.

The vector projection of [latex]\mathbf{v}[/latex] onto [latex]\mathbf{u}[/latex], written as [latex]\text{proj}_{\mathbf{u}}\mathbf{v}[/latex], represents the component of [latex]\mathbf{v}[/latex] that acts in the direction of [latex]\mathbf{u}[/latex].

This image has a vector labeled “v.” There is also a vector with the same initial point labeled “proj sub u v.” The third vector is from the terminal point of proj sub u v in the same direction labeled “u.” A broken line segment from the initial point of u to the terminal point of v is drawn and is perpendicular to u.
Figure 2. The projection of [latex]{\bf{v}}[/latex] onto [latex]{\bf{u}}[/latex] shows the component of vector [latex]{\bf{v}}[/latex] in the direction of [latex]{\bf{u}}[/latex].

From the geometry of the situation, if [latex]\theta[/latex] is the angle between [latex]\mathbf{u}[/latex] and [latex]\mathbf{v}[/latex], then the length of the projection is:

[latex]||\text{proj}_{\mathbf{u}}\mathbf{v}|| = ||\mathbf{v}|| \cos \theta[/latex]

Using the dot product formula [latex]\cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{||\mathbf{u}|| , ||\mathbf{v}||}[/latex], this becomes:

[latex]||\text{proj}_{\mathbf{u}}\mathbf{v}|| = ||\mathbf{v}|| \cdot \frac{|\mathbf{u} \cdot \mathbf{v}|}{||\mathbf{u}|| , ||\mathbf{v}||} = \frac{|\mathbf{u} \cdot \mathbf{v}|}{||\mathbf{u}||}[/latex]

To get the actual projection vector (not just its length), we multiply by a unit vector in the direction of [latex]\mathbf{u}[/latex]:

[latex]||\text{proj}_{\mathbf{u}}\mathbf{v}|| = \frac{\mathbf{u}\cdot\mathbf{v}}{||\mathbf{u}||}(\frac{1}{||\mathbf{u}||}\mathbf{u}) = \frac{\mathbf{u} \cdot \mathbf{v}}{||\mathbf{u}||^2}\mathbf{u}[/latex].

The length of this vector is also known as the scalar projection of [latex]\mathbf{v}[/latex] onto [latex]\mathbf{u}[/latex] and is denoted by

[latex]||\text{proj}_\text{u}\mathbf{v}|| = \text{comp}_\text{u}\mathbf{v} = \frac{\mathbf{u}\cdot\mathbf{v}}{||\mathbf{u}||}[/latex].

vector projection

Vector Projection Formula

[latex]\text{proj}_{\mathbf{u}}\mathbf{v} = \frac{\mathbf{u} \cdot \mathbf{v}}{||\mathbf{u}||^2}\mathbf{u}[/latex]

Scalar Projection (length only)

[latex]||\text{proj}_{\mathbf{u}}\mathbf{v}|| = \text{comp}_{\mathbf{u}}\mathbf{v} = \frac{\mathbf{u} \cdot \mathbf{v}}{||\mathbf{u}||}[/latex]

Think of projection as the “shadow” that vector [latex]\mathbf{v}[/latex] casts onto vector [latex]\mathbf{u}[/latex] when light shines perpendicular to [latex]\mathbf{u}[/latex].

Find the projection of [latex]\mathbf{v}[/latex] onto [latex]\mathbf{u}[/latex].

  1. [latex]\mathbf{v} = \langle 3,5,1 \rangle[/latex] and [latex]\mathbf{u} = \langle -1,4,3 \rangle[/latex]
  2. [latex]\mathbf{v} = 3\mathbf{i} -2\mathbf{j}[/latex] and [latex]\mathbf{u} = \mathbf{i} +6\mathbf{j}[/latex]

Decomposing Vectors into Orthogonal Components

Projections allow us to decompose (break apart) vectors into orthogonal components. This process is called resolution of a vector into components.

Let [latex]\mathbf{v} = \langle 6,-4 \rangle[/latex] and [latex]\mathbf{u} = \langle 3,1 \rangle[/latex]. Decompose [latex]\mathbf{v}[/latex] into orthogonal components where one component has the same direction as [latex]\mathbf{u}[/latex].

Step 1: Find the projection of [latex]\mathbf{v}[/latex] onto [latex]\mathbf{u}[/latex]: [latex]\mathbf{p} = \text{proj}_{\mathbf{u}}\mathbf{v} = \frac{\mathbf{u} \cdot \mathbf{v}}{||\mathbf{u}||^2}\mathbf{u}[/latex]

[latex]\mathbf{u} \cdot \mathbf{v} = (3)(6) + (1)(-4) = 18 - 4 = 14[/latex]

[latex]||\mathbf{u}||^2 = 3^2 + 1^2 = 10[/latex]

[latex]\mathbf{p} = \frac{14}{10}\langle 3,1 \rangle = \frac{7}{5}\langle 3,1 \rangle = \left\langle \frac{21}{5}, \frac{7}{5} \right\rangle[/latex]

Step 2: Find the orthogonal component:

[latex]\mathbf{q} = \mathbf{v} - \mathbf{p} = \langle 6,-4 \rangle - \left\langle \frac{21}{5}, \frac{7}{5} \right\rangle = \left\langle \frac{9}{5}, -\frac{27}{5} \right\rangle[/latex]

Step 3: Verify orthogonality:

[latex]\mathbf{q} \cdot \mathbf{p} = \left\langle \frac{9}{5}, -\frac{27}{5} \right\rangle \cdot \left\langle \frac{21}{5}, \frac{7}{5} \right\rangle = \frac{189}{25} - \frac{189}{25} = 0[/latex] ✓

Therefore: [latex]\mathbf{v} = \mathbf{p} + \mathbf{q}[/latex] where [latex]\mathbf{p}[/latex] and [latex]\mathbf{q}[/latex] are orthogonal.

vector decomposition

Any vector [latex]\mathbf{v}[/latex] can be written as [latex]\mathbf{v} = \mathbf{p} + \mathbf{q}[/latex] where:

  • [latex]\mathbf{p} = \text{proj}_{\mathbf{u}}\mathbf{v}[/latex] (component parallel to [latex]\mathbf{u}[/latex])
  • [latex]\mathbf{q} = \mathbf{v} - \mathbf{p}[/latex] (component orthogonal to [latex]\mathbf{u}[/latex])

This decomposition is unique and [latex]\mathbf{p} \cdot \mathbf{q} = 0[/latex].

Express [latex]\mathbf{v} = \langle 8,-3,-3 \rangle[/latex] as a sum of orthogonal vectors such that one of the vectors has the same direction as [latex]\mathbf{u} = \langle 2,3,2 \rangle[/latex].

A container ship leaves port traveling [latex]15^\circ[/latex] north of east. Its engine generates a speed of [latex]20[/latex] knots along that path (see the following figure). In addition, the ocean current moves the ship northeast at a speed of [latex]2[/latex] knots. Considering both the engine and the current, how fast is the ship moving in the direction [latex]15^\circ[/latex] north of east? Round the answer to two decimal places.

This figure is an image of a ship. The ship is at the origin of two perpendicular axes. The horizontal axis is labeled “east.” The second axis is vertical and labeled “north.” From the ship there are two vectors. The first is labeled “v” and has an angle of 15 degrees between the East axis and the vector v. The second vector is labeled “w” and has an angle of 45 degrees between the East axis and the vector w.
Figure 3. Find how fast the ship is moving in the direction of [latex]15^{\circ}[/latex] north of east.