## Projection of Two Vectors

The **dot product** can be used to perform a projection.

The vector projection of a vector onto a non-zero vector results in a straight line which is parallel to , which includes a vector *orthogonal* to .

I've done my best to show this visualization here:

*Note to self: Learn how to use some graphing software, or something. ðŸ˜„*

We can think of the projection of onto as the **shadow** that casts on when rays of light are perpendicular to .

Going back to the diagram, we want to know:

Where does intersect with so that can be a perpendicular to ?

- As in, what is the displacement of from ?

We also want to know where meets with .

We'll need to find 2 things:

- The vector
- The vector

We can begin to find by using this formula:

Then, we can say that is the projection of onto .

**Note**: If is orthogonal to , then .

### Step-By-Step Process

Here is the step-by-step process in finding projection of two vectors.

#### Step 1 - Find the Length of

Back to that diagram above the page, we can find the length of using trigonometry. Assume that we know :

#### Step 2 - Normalize , and

Knowing that for unit vectors , and , the dot products of these unit vectors is:

Then, we know that finding the unit vector of to be , and the unit vector of to be , we can write an expression like this:

Since evaluate to due to them being unit vectors, we can simplify to have the formula be written without it.

#### Step 3 - Rewrite

We can now rewrite the equation for from to:

We read this as two components:

- is a scalar, and will adjust to be the required length. It is the .
- will indicate the result to be in the same orientation as .

#### Step 4 - Find

Now, we can find the orthogonal vector to . This will be the vector . We can rewrite our expression to be

Since w know , and now , the process then becomes an exercise of vector subtraction.

### Example

I ripped this one off from here.

Suppose we have two vectors:

We now want to find the projection of onto . In order to do that, we have to:

- Find
- Find .

First let's normalize the vectors , and to unit vectors.

And, for :

Now, we find by

Now, it becomes a scalar multiplication:

So, we know that the displacement .

Next, let's find .

So, , meaning that this is the displacement starting from . If we plot this, we will find that is orthogonal to .

Now, excuse the crappy picture

### Notes

It is important to know that projected onto is

**NOT**the same as projected onto .The vector resulting from a projection is oriented in the same direction as the vector projected onto.

Since the cosine function is never more than , the length of the projection will

**NEVER**be longer than the projected vector.