# vector dot product question

• Feb 2nd 2011, 12:23 PM
duaneg37
vector dot product question
If I define vectors $\vec{CA}=\vec{V}+\vec{U}$ and $\vec{CB}=\vec{V}-\vec{U}$ is the following true?

$\vec{CA}\cdot\vec{CB}=\vec{V}^{2}-\vec{U}^{2}$

thanks for the help!
• Feb 2nd 2011, 12:26 PM
Ackbeet
The dot product obeys the usual foiling rule for multiplying out. In addition, the dot product is commutative. Those two facts should enable you to tell me the answer to your question.
• Feb 2nd 2011, 12:34 PM
duaneg37
I thought the dot product gave a number, not a vector. So I am not sure.
• Feb 2nd 2011, 12:36 PM
Ackbeet
Quote:

Originally Posted by duaneg37
I thought the dot product gave a number, not a vector. So I am not sure.

It does. But the square of a vector (the square of its length) is also a number. So there at least a fighting chance that you're right. Show me what you get when you foil out the dot product in the usual way.
• Feb 2nd 2011, 12:42 PM
Plato
You are correct. The dot product is a scalar.
I think your mistake is in notation.
$\vec{CA}\cdot\vec{CB}=\vec{V}\cdot\vec{V}-\vec{U}\cdot\vec{U}$. Do see why that is the case?
Now $\vec{V}\cdot\vec{V}=\|\vec{V}\|^2$, the length squared.
So we have $\|\vec{V}\|^2-\|\vec{U}\|^2$, a number.
• Feb 2nd 2011, 12:44 PM
Ackbeet

Perhaps. But I've seen notation abused in this manner all the time, especially by physicists.
• Feb 2nd 2011, 12:47 PM
Plato
Quote:

Originally Posted by Ackbeet
Perhaps. But I've seen notation abused in this manner all the time, especially by physicists.

Suppose that $AB$ is the diameter of a circle with center $O$, and that $C$ is a point on one of the two arcs joining $A$ and $B$. Show that $\vec{CA}$ and $\vec{CB}$ are orthogonal. $\vec{V}$ is from $O$ to $C$. $\vec{U}$ is from $O$ to $B$. $\vec{-U}$ is from $O$ to $A$.
I know if their dot product is zero, they are orthogonal. If the dot product is $\vec{U}^{2}-\vec{V}^{2}$ it would have to be zero because their magnitudes are the same.