# Is it permissible to factor out a dot product?

I am an independent student trying to work out a proof. I have followed steps to get to a case where I have $c \vec v \cdot \vec y > d\vec v \cdot \vec x$ -- where c and d are scalars, and v,y and x are vectors. In simple algebra I would factor out the v's -- but I am still learning linear algebra. Is there an analogous operation? It seems like vectors cannot be divided like scalars.

• Just to be clear; $c$ and $d$ are scalars, $\bf v$, $\bf x$ and $\bf y$ are vectors and $>$ is an inequality sign? It's not true that you can "cancel out" the $v$'s, but what you can do is say $c {\bf v} \cdot {\bf y} > d {\bf v} \cdot {\bf x} \iff {\bf v} \cdot (c{\bf y} - d{\bf x}) > 0$. Try to work with the bilinearity of the dot product. Jun 17, 2015 at 14:44
• What ultimately would you want to prove if you could make some statement about $y$ relative to $x$? Jun 17, 2015 at 14:46

But in the above you can use that $$c(\vec{v}\cdot\vec{y}) = (c\vec{v})\cdot \vec{y} = \vec{v}\cdot(c\vec{y}).$$ That is, you can move the scalar around.
There are other rules, as $$\vec{v}\cdot \vec{y} + \vec{v}\cdot\vec{z} = \vec{v}\cdot (\vec{y} + \vec{z}).$$ Here you are factoring out the vector $\vec{v}$.
• @bernie2436: Yes, if $V$, $X$ and $Z$ are vectors and $d$ is a scalar, you can do this. Jun 17, 2015 at 15:17