# Justin's Blog

## February 19, 2012

### Vectors came from Quaternions

Before vectors, there were quaternions. Quaternions are the next step after complex numbers from real numbers.

A complex number may be written as $$z = x + \hat{i} y$$ Where $$\hat{i} = - 1$$

A quaternion may be written as $$q = w + \hat{i}x + \hat{j} y + \hat{k}z$$

w is the scalar part, and the rest is the vector part.

If you don't know how to multiply i, j, and k, please learn how before continuing.

### On to Vectors

A vector is just a quaternion where the scalar part is zero (w=0). $$\vec{v}= x \hat{i}+ y\hat{j} + z\hat{k}$$

The basis vectors are $$\hat{i}, \hat{j}, \hat{k}$$

You can also write a vector like this: $$\vec{v}= (x,y,z)$$

### Points in ℝ 3

So, a vector can be written as $$\vec{v}= (x,y,z)$$

But a point can also be written as $$P = (x,y,z)$$

This may seem ambiguous at first, but it isn't. A point is a special vector that describes a distance from the origin.

### Distance Between Points

A vector can represent a distance between points.

All that is needed to find this distance is to subtract one point from another.

So if

$$P = (3,4,5)$$ $$Q= (0,0,0)$$

Then the vector from Q to P is $$\vec{QP} = \vec{P} - \vec{Q}$$ $$\vec{QP} = (3, 4, 5)$$

Remember, subtract the start from the end to get the distance. This might seem backwards, but think about it.

If you started at work, 3 miles from home and got lunch 4 miles from home in the same direction, you would have moved 1 mile away from home.

That's because

(end)-(start)=(distance)

4 - 3 = 1

On your way back from lunch to work, you start 4 miles away and end 3 miles away. You moved -1 mile away from home (1 mile toward home).

Another example: $$P = (3, 4, 5)$$ $$Q = (-1, 7, 3)$$ $$\vec{QP} = (4, -3, 2)$$

When adding and subtracting vectors, just add or subtract each component to create a new vector.

Adding two vectors is like putting the tail of one vector at the head of another and finding vector that makes a straight path from the beginning of the first vector to the end of the second.

Subtracting two vectors is like adding the negative of the second vector to the first vector.

### Multiplying Vectors

Let's multiply two vectors in algebraic form, using the distributive property.

$$\vec{v}_{1} = x_{1}\hat{i} + y_{1}\hat{j} + z_{1}\hat{k}$$

$$\vec{v}_{2}=x_{2}\hat{i}+y_{2}\hat{j}+z_{2}\hat{k}$$

$$\vec{v}_{1} \cdot \vec{v}_{2} = x_{1}\hat{i}(x_{2}\hat{i}+y_{2}\hat{j}+z_{2}\hat{k}) +$$

$$y_{1}\hat{j}(x_{2}\hat{i}+y_{2}\hat{j}+z_{2}\hat{k}) +$$

$$z_{1}\hat{k}(x_{2}\hat{i}+y_{2}\hat{j}+z_{2}\hat{k})$$

We can group these terms into squared and mixed basis vectors.

$$\vec{v}_{1} \cdot \vec{v}_{2} = x_{1}x_{2}\hat{i}^{2}+y_{1}y_{2}\hat{j}^{2}+z_{1}z_{2}\hat{k}^{2}+$$

$$y_{1}z_{2}\hat{j}\hat{k}+z_{1}y_{2}\hat{k}\hat{j}+ x_{1}z_{2}\hat{i}\hat{k}+$$

$$z_{1}x_{2}\hat{k}\hat{i}+ x_{1}y_{2}\hat{i}\hat{j}+y_{1}x_{2}\hat{j}\hat{i}$$

You can count these terms to make sure I didn't miss any. There should be 9.

The first three terms (in the parentheses) are called the dot product. The rest are the cross-product.

See how the cross product is the product of the cross terms. The dot product is the ... well... dot the i's cross the t's you know...

I don't know why it's called the dot product. Anyway, we can simplify this a bit.

Remember the rules for multiplying i, j, and k?

$$\vec{v}_{1}*\vec{v}_{2}= -(x_{1}x_{2}+y_{1}y_{2}+z_{1}z_{2})+$$

$$(y_{1}z_{2}-z_{1}y_{2})\hat{i}+$$

$$-(x_{1}z_{2}+z_{1}x_{1})\hat{j}+$$

$$(x_{1}y_{2}-y_{1}x_{2})\hat{k}$$

The first term is the dot product. Notice how it has no basis vectors? It is a scalar.

The cross product is the vector part. That is, it is the sum of the terms with basis vectors.