# dot product identity

$$a \cdot (a \cdot b)=(a \cdot a)(a \cdot b)$$

Is this identity true when $a$ and $b$ are vectors, and when $\cdot$ is the dot product operator? And assuming that $()()$ means multiplying the contents of the parentheses.

Can anyone please post explanation and links of this identity? I saw it used somewhere, but I cannot seem to find documentation of it in textbooks or in search engine research using the key words I could think of.

Thank you.

It is used in a suggested solution for the following problem:

"Show that the vector orth_a b = b - proj_a b is orthogonal to a. (It is called an orthogonal projection of b.)"

Where a and b are vectors, and proj_a b is the vector projection of b onto a.

Really what I need is to learn to answer this problem, and the part of the suggested proof I don't understand is given in my initial post above. If you have a better answer to this problem, I would love to learn how your approach works.

-
Note that $a$ is a vector and $a\cdot b$ is a scalar, so $a\cdot (a\cdot b)$ is not defined. Perhaps you mean something else? – Alex Becker Sep 7 '12 at 22:49
@AlexBecker, Thank you. Please see my edit of my original post above. It is placed under the dividing line under thank you. – CodeMed Sep 7 '12 at 22:59

When you have two vectors $\vec{a}$ and $\vec{b}$ you can take their dot product: $\vec{a}\cdot\vec{b}$. This dot product is a scalar (number). It is indeed sometimes called the scalar product.

It does not make sense to take a dot product of a vector with a scalar, so what you have written on the left hand side is not well defined (since here you have the dot-product of a vector $\vec{a}$ and a scalar $(\vec{a}\cdot\vec{b})$.

If you want, you can take a look at the Wikipedia article on the dot product. Under properties, you can find a few formulas. Try to go through them and convince yourself that they are well defined (i.e. that you only have dot products between vectors - noting of course that one can multiply a vector by a scalar)

Edit: Just to add a bit more about your original problem. You have that $$proj_\vec{a}(\vec{b}) = \frac{\vec{a}\cdot \vec{b}}{\lvert \vec{a} \lvert^2}\vec{a}$$ and $$orth_\vec{a}(\vec{b}) = \vec{b} - proj_\vec{a}(\vec{b}).$$ So show that $\vec{a}$ and $orth_\vec{a}(\vec{b})$ are orthogonal you conpute their dot product: \begin{align} \vec{a}\cdot (\vec{b} - proj_\vec{a}(\vec{b})) &= \vec{a} \cdot \vec{b} - \vec{a}\cdot proj_\vec{a}(\vec{b}) \\ &= \vec{a} \cdot \vec{b} - \vec{a}\cdot \frac{\vec{a}\cdot \vec{b}}{\lvert \vec{a} \lvert^2}\vec{a} \\ &= \vec{a} \cdot \vec{b} - \frac{\vec{a}\cdot \vec{b}}{\lvert \vec{a} \lvert^2}\vec{a}\cdot \vec{a} \\ &= \vec{a} \cdot \vec{b} - \frac{\vec{a}\cdot \vec{b}}{\lvert \vec{a} \lvert^2}\lvert \vec{a}\vert^2 \\ &= \dots \end{align}

You can probably finish it from here.

-
Thank you. Please see my edit to my original post above. I did look at the wikipedia posting. This is really an attempt to answer the broader problem that I added in my edit above. I think that the suggested proof I was given might not be right. I would love to hear your suggestions as to how to better show the result requested in my edit to my original posting above. Thank you. – CodeMed Sep 7 '12 at 23:02
@CodeMed: did my edit help? – Thomas Sep 7 '12 at 23:12
Thank you. Actually, the very next step at the end of your edit is the part about which I posted my initial question above: a.a.b=? Can you elaborate the next couple steps? – CodeMed Sep 7 '12 at 23:36
@CodeMed: sure, I will edit – Thomas Sep 7 '12 at 23:43
@CodeMed: Did that help? – Thomas Sep 7 '12 at 23:45