If I define vectors $\displaystyle \vec{CA}=\vec{V}+\vec{U}$ and $\displaystyle \vec{CB}=\vec{V}-\vec{U}$ is the following true?

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

thanks for the help!

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- Feb 2nd 2011, 12:23 PMduaneg37vector dot product question
If I define vectors $\displaystyle \vec{CA}=\vec{V}+\vec{U}$ and $\displaystyle \vec{CB}=\vec{V}-\vec{U}$ is the following true?

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

thanks for the help! - Feb 2nd 2011, 12:26 PMAckbeet
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 PMduaneg37
I thought the dot product gave a number, not a vector. So I am not sure.

- Feb 2nd 2011, 12:36 PMAckbeet
- Feb 2nd 2011, 12:42 PMPlato
You are correct. The dot product is a scalar.

I think your mistake is in notation.

$\displaystyle \vec{CA}\cdot\vec{CB}=\vec{V}\cdot\vec{V}-\vec{U}\cdot\vec{U}$. Do see why that is the case?

Now $\displaystyle \vec{V}\cdot\vec{V}=\|\vec{V}\|^2$, the length squared.

So we have $\displaystyle \|\vec{V}\|^2-\|\vec{U}\|^2$, a number. - Feb 2nd 2011, 12:44 PMAckbeet
Reply to Plato:

Perhaps. But I've seen notation abused in this manner all the time, especially by physicists. - Feb 2nd 2011, 12:47 PMPlato
- Feb 2nd 2011, 01:01 PMduaneg37
Suppose that $\displaystyle AB$ is the diameter of a circle with center $\displaystyle O$, and that $\displaystyle C$ is a point on one of the two arcs joining $\displaystyle A$ and $\displaystyle B$. Show that $\displaystyle \vec{CA}$ and $\displaystyle \vec{CB}$ are orthogonal. $\displaystyle \vec{V}$ is from $\displaystyle O$ to $\displaystyle C$. $\displaystyle \vec{U}$ is from $\displaystyle O$ to $\displaystyle B$. $\displaystyle \vec{-U}$ is from $\displaystyle O$ to $\displaystyle A$.

I know if their dot product is zero, they are orthogonal. If the dot product is $\displaystyle \vec{U}^{2}-\vec{V}^{2}$ it would have to be zero because their magnitudes are the same.

Is this enough proof? I think it is. - Feb 2nd 2011, 01:06 PMAckbeet
Had to think about it for a while, but yes, I think your proof works fine.