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### Crossed up by Scalars and Vectors

```Date: 02/14/2012 at 06:21:31
From: Karthik

I have never really understood why dot products and cross products have
been defined the way they are. Why is the result of a dot product a
scalar, and that of a vector product a vector?

I have consulted many books, and I have searched the net, but nobody seems
to give any plausible explanation for this.

```

```
Date: 02/14/2012 at 08:49:17
From: Doctor Jerry

Hello Karthik,

Thanks for writing to Dr. Math.

Given the idea of a vector, it is natural to look for a way to calculate
the angle t between vectors {a1, a2, a3} and {b1, b2, b3}. Letting ||a||
be the length of the vector a, if you apply the Law of Cosines to the
triangle with sides a, b, and a - b, you will see

(a1 - b1)^2 + (a2 - b2)^2 + (a3 - b3)^2 =
a1^2 + a2^2 + a3^2 + b1^2 + b2^2 + b3^2 - 2||a||*||b||* cos[t]

Simplify and you will find this, where "x y" means "x times y":

a1 b1 + a2 b2 + a3 b3 = ||a||*||b||*cos[t]

This makes it clear that the left side is a useful combination of a and b.

For the cross product, if you have vectors a = {a1, a2, a3} and
b = {b1, b2, b3}, it seems clear that a vector that is perpendicular to
both of a and b would be useful. So, we seek a vector {c1, c2, c3}
such that

a.c = 0
b.c = 0

If you solve this system of two equations for c1 and c2, you will find

c1 = -(-a3 b2 c3 + a2 b3 c3)/(a2 b1 - a1 b2)
c2 = -(a3 b1 c3 - a1 b3 c3)/(a2 b1 - a1 b2)

If you choose c3 as a2 b1 - a1 b2, this will give a solution and clean up
the mess a bit.

Please feel free to write back -- using the URLs at the bottom of this
message -- if you have questions relative to my comments.

- Doctor Jerry, The Math Forum
http://mathforum.org/dr.math/

```

```
Date: 02/16/2012 at 03:07:00
From: Karthik

But could you please explain to me why the result of a dot product is a
scalar and that of a cross product, a vector?

```

```
Date: 02/16/2012 at 07:20:48
From: Doctor Jerry

Hello,

Thanks for writing to Dr. Math.

1. I said:

"Given the idea of a vector, it is natural to look for a way to
calculate the angle t between vectors {a1, a2, a3} and {b1, b2, b3}."

The angle t or its cosine are scalars.

2. I also said:

"For the cross product, if you have vectors a = {a1, a2, a3} and
b = {b1, b2, b3}, it seems clear that a vector that is perpendicular
to both...."

The search starts with the idea of scalar and vector in these two situations.

Please feel free to write back -- using the URLs at the bottom of this
message -- if you have questions relative to my comments.

- Doctor Jerry, The Math Forum
http://mathforum.org/dr.math/

```

```
Date: 02/16/2012 at 11:35:29
From: Karthik
Subject: Thank you (Confusion about vectors.)

Thank you very much for all the help!
```
Associated Topics:
High School Linear Algebra

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