Vector Projections

# Vector Projections

 Definition: Let $\vec{u}, \vec{v} \in \mathbb{R}^n$. Then the Vector Projection of $\vec{u}$ Along $\vec{v}$ denoted $\vec{u}_\vec{v} = \mathrm{proj}_{\vec{v}} \vec{u} = \frac{\vec{u} \cdot \vec{v}}{\| \vec{v} \|^2} \cdot \vec{v}$ is a scalar multiple of $\vec{v}$ such that for some vector $\vec{w} \in \mathbb{R}^n$ such that $\mathrm{proj}_{\vec{v}} \vec{u} \perp \vec{w}$ we have that $\mathrm{proj}_{\vec{v}} \vec{u} + \vec{w} = \vec{u}$. The signed length of $\mathrm{proj}_{\vec{v}} \vec{u}$ is the Scalar Projection of $\vec{u}$ Along $\vec{v}$ denoted $\mathrm{comp}_{\vec{v}} \vec{u} = \frac{ \vec{u} \cdot \vec{v} }{\| \vec{v} \|}$.

Before we look at some examples of vector projections, we will first verify the formulas $\mathrm{proj}_{\vec{v}} \vec{u} = \frac{\vec{u} \cdot \vec{v}}{\| \vec{v} \|^2} \cdot \vec{v}$ and $\mathrm{comp}_{\vec{v}} \vec{u} = \frac{ \vec{u} \cdot \vec{v} }{\| \vec{v} \|}$ provided from the definition.

Let $\vec{u}, \vec{v}, \vec{w}, \vec{u}_{\vec{v}} \in \mathbb{R}^n$, and suppose that $\mathrm{proj}_{\vec{v}} \vec{u} = \vec{u}_{\vec{v}}$ runs along $\vec{v}$. Then $\vec{u}_{\vec{v}} \| \vec{v}$ and so for some scalar $k$ we have that $\vec{u}_{\vec{v}} = k \vec{v}$. We also have that $\vec{u} = \vec{u}_{\vec{v}} + \vec{w}$ where $\vec{u}_{\vec{v}} \perp \vec{w}$. Substituting $\vec{u}_{\vec{v}} = k \vec{v}$ we get that $\vec{u} = k\vec{v} + \vec{w}$. Now we will take the dot product of both sides of this equation with $\vec{v}$ and noting that $\vec{w} \cdot \vec{v} = 0$ since $\vec{w} \perp \vec{v}$ we have that:

(1)
\begin{align} \vec{u} \cdot \vec{v} = (k \vec{v} + \vec{w}) \cdot \vec{v} \\ \vec{u} \cdot \vec{v} = (k \vec{v} \cdot \vec{v}) + (\vec{w} \cdot \vec{v}) \\ \vec{u} \cdot \vec{v} = k \vec{v} \cdot \vec{v} \\ \vec{u} \cdot \vec{v} = k \| \vec{v} \| ^2 \\ k = \frac{\vec{u} \cdot \vec{v}}{\| \vec{v} \|^2 } \end{align}

Therefore since $\vec{u}_{\vec{v}} = k \vec{v}$ we have that $\vec{u}_{\vec{v}} = \mathrm{proj}_{\vec{v}} \vec{u} = \frac{\vec{u} \cdot \vec{v}}{ \| \vec{v} \|^2 } \vec{v}$.

Now to find the scalar component of $\vec{u}$ along $\vec{v}$, we first need to take the magnitude of the projection vector.

(2)
\begin{align} \quad \biggr \| \mathrm{proj}_{\vec{v}} \vec{u} \biggr \| = \biggr \| \frac{\vec{u} \cdot \vec{v}}{\| \vec{v} \|^2} \vec{v} \biggr \| = \frac{\| \vec{u} \cdot \vec{v} \|}{\| \vec{v} \|^2} \| \vec{v} \| = \frac{ \mid \vec{u} \cdot \vec{v} \mid }{\| \vec{v} \|} \end{align}

Since $\mathrm{comp}_{\vec{v}} \vec{u}$ is the signed length/magnitude of the projection vector, we can remove the absolute value bars so that we then have that $\mathrm{comp}_{\vec{v}} \vec{u} = \frac{\vec{u} \cdot \vec{v}}{\| \vec{v} \|}$.

Now let's look at some examples regarding vector projections.

## Example 1

Let $\vec{u} = (1, 4)$ and $\vec{v} = (2, 5)$. Determine $\mathrm{proj}_{\vec{v}} \vec{u}$.

To apply the formula we derived above, we will need to first calculate $\vec{u} \cdot \vec{v}$ and $\| \vec{v} \|^2$. First, $\vec{u} \cdot \vec{v} = (1, 4) \cdot (2, 5) = (1)(2) + (4)(5) = 22$. Now $\| \vec{v} \|^2 = (\sqrt{2^2 + 5^2})^2 = 29$. Therefore:

(3)
\begin{align} \mathrm{proj}_{\vec{v}} \vec{u} = \frac{\vec{u} \cdot \vec{v}}{\| \vec{v} \|^2} \vec{v} \\ \mathrm{proj}_{\vec{v}} \vec{u} = \frac{22}{29} (2, 5) \\ \mathrm{proj}_{\vec{v}} \vec{u} = \left ( \frac{44}{29}, \frac{110}{29} \right ) \end{align}

## Example 2

Let $\vec{u} = (-1, 1, 6)$ and $\vec{v} = (-2, 0, 3)$. Determine $\mathrm{proj}_{\vec{v}} \vec{u}$.

Once again, let's compute $\vec{u} \cdot \vec{v}$ and $\| \vec{v} \| ^2$ again. First, $\vec{u} \cdot \vec{v} = (-1, 1, 6) \cdot (-2, 0, 3) = (-1)(-2) + (1)(0) + (6)(3) = 2 + 0 + 18 = 20$. Now $\| \vec{v} \| ^2 = ( \sqrt{(-2)^2 + 0^2 + 3^2})^2 = 4 + 9 = 13$. Therefore:

(4)
\begin{align} \mathrm{proj}_{\vec{v}} \vec{u} = \frac{\vec{u} \cdot \vec{v}}{\| \vec{v} \|^2} \vec{v} \\ \mathrm{proj}_{\vec{v}} \vec{u} = \frac{20}{13} (-2, 0, 3) \\ \mathrm{proj}_{\vec{v}} \vec{u} = \left ( \frac{-40}{13}, 0, \frac{60}{13} \right ) \end{align}