The dot product of two vectors A and B is a key operation in using vectors in geometry.

In the coordinate space of any dimension (we will be mostly interested in dimension 2 or 3):

**Definition**: If A = (a_{1}, a_{2}, ..., a_{n})
and B = (b_{1}, b_{2}, ..., b_{n}), then the **dot
product** A^{.}B = a_{1}b_{1}
+ a_{2} b_{2} + ... + a_{n} b_{n}

**Examples:** Let A = (1, 2, -1), B = (3, 2, 1), C = (0, -5, 2). Note that
if we set D = 2B - C, then D = (6, 9, 0).

A^{.}B = 1*3 + 2*2 + (-1)*1 = 6

A^{.}C = 1*0 + 2*(-5) + (-1)*2 = -12

A^{.}D = 1*6 + 2*9 + (-1)*0 = 24

Notice that A^{.}D = A^{.}(2B - C) = 24 and also 2A^{.}B
- A^{.}C = 2*6 - (-12) = 24. This is not a coincidence, but follows
from the **algebraic properties** of the dot product..

These properties are extremely important, though they are a little boring to prove. It takes a second look to see that anything is going on at all, but look twice or 3 times.

(1) **(Commutative Property)** For any two vectors A and B, A^{.}B
= B^{.}A.

(2)** (Scalar Multiplication Property)** For any two vectors A and B and
any real number c, (cA)^{.}B = A^{.}(cB) = c(A^{.}B)

(3) **(Distributive Property)** For any 3 vectors A, B and C, A^{.}(B+C)
= A^{.}B + A^{.}C.

Let A, B, C, D be as above for the next 3 exercises.

**Exercise 1**: Compute B^{.}A and compare with A^{.}B.
Can you see why these numbers are the same in this example and will always be
the same for any choice of A and B?

**Exercise 2**: Let c = 10. Write down 10A and 10B. Then compute each of
the 3 terms in property (2) above and see that they are in fact the same. Again
look to see why this is true.

**Exercise 3**: Compute E = B+C. Check that A^{.}E is really the
sum of A^{.}B and A^{.}C.

**Exercise 4**. Show how you can combine (3) with (2) in several steps to
show that for **any** vectors A, B, C and any real numbers h and k, A^{.}(hB+kC)
= A^{.}hB + A^{.}kC. Tell how this explains why for the special
A, B, C, D above, A^{.}D = A^{.}(2B - C).

**Exercise 5: **Use the properties above to expand (aA + bC)^{.}(cC
+ dD) into a sum of four terms, starting with ac(A^{.}C) + ...

**Exercise 6:** For a point A = (3, 4) in the plane, compute the square
root of A^{.}A. Tell why this number is the distance from A to O.

For A = (a_{1}, a_{2}, ..., a_{n}), the dot product
A^{.}A is simply the sum of squares of each entry.

In the plane or 3-space, the Pythagorean theorem tells us that the distance from O to A, which we think of as the length of vector OA, (or just length of A), is the square root of this number.

**Definition**. For A in n-space, the length of A = square root of A^{.}A.
This length is written |A|. so |A|^{2} = A^{.}A.

Likewise, the Pythagorean theorem also shows that the distance from A to B is the length of AB, which is the length of B-A.

**Definition**: For A and B in n-space, the distance from A to B is the
length |B-A|.

Note: This equals |A-B| also.

**Cosine Theorem: **For A and B in the plane or in space, A^{.}B
= |A| |B| cos AOB

In the triangle AOB, if we let the length of side AB = c, then by definition of distance,

c^{2}= |A - B|^{2}

Proof: From the algebraic properties,

|A - B|^{2} = A^{.}B - B^{.}A - B^{.}B
= |A|^{2} + |B|^{2} - 2A^{.}B.

But from geometry, since |A| = |OA| = side opposite B, etc., the Law of Cosines for triangle AOB says

c^{2}= |A|^{2} + |B|^{2} - 2|A| |B|
cos AOB.

Comparing these, all but one term is the same in each, so we see that A^{.}B
= |A| |B| cos AOB

**Exercise:** Compute the angle between (1, 1, 1) and (0, 0, 1).

**Exercise:** In the plane, let A = (1, 2), B = (3, 4), C = (-2, -1). Use
the dot product to compute all the side lengths and all the angles of this triangle.

**Exercise:** In the plane, let A = (1, 2, 1), B = (3, 4, 1), C = (-2, -1,
3). Use the dot product to compute all the side lengths and all the angles of
this triangle.

The cosine of a right angle = 0, so a very important special case of the cosine theorem is this:

**Orthogonal Vector Theorem: **Two vectors A and B are orhthogonal if and
only if their dot product is zero.

Dot product definitions and examples at Texas A&M