# Dot product equation

Given a constant vector $\mathbf{c}$ and a vector of variables $\mathbf{x}$, what can we say about the equation:

$\langle\mathbf{c},\mathbf{x}\rangle=s$

where $\langle*,*\rangle$ represents the dot product and $s$ is a real number.

This is a sum of $c_ix_i$ terms and since we have only one equations and more unknowns we have an infinity of solutions. However, how can I say anything about this equation?

I tried to use the representation of the dot product $\langle\mathbf{c},\mathbf{x}\rangle=\|\mathbf{c}\|\|\mathbf{x}\|\cos\theta$, where $\theta$ is the angle between the vectors.

$\langle\mathbf{c},\mathbf{x}\rangle=\|\mathbf{c}\|\|\mathbf{x}\|\cos\theta=s$ $\implies \|\mathbf{x}\|=\frac{s}{\|\mathbf{c}\|\cos\theta}$

Is this the most we can say about this equation and its "solution"? For example, we can see that $\|\mathbf{x}\|$ will be in the interval $[\frac{s}{\|\mathbf{c}\|},\frac{s}{\|\mathbf{c}\|}]$.

What else? Can we, for example, find a bound for $\langle\mathbf{c},\mathbf{x}\rangle$ ?

• What are you looking for? – Alex Silva Dec 2 '14 at 23:48
• If the equation is $\langle\bf{c, x}\rangle = s$ then I'd say the bound for $\langle\bf{c, x}\rangle$ is $s$. – Jim Dec 3 '14 at 0:00

Thinking geometrically, $\langle\mathbf{c},\mathbf{x}\rangle=s$ is equivalent to

$$\langle\frac{\mathbf{c}}{||\mathbf{c}||},\mathbf{x}\rangle= \frac{s}{||\mathbf{c}||}$$

This then is satisfied by all vectors $\mathbf{x}$ that have a projection of $\displaystyle \frac{s}{||\mathbf{c}||}$ in direction $\displaystyle \hat{\mathbf{c}} = \frac{\mathbf{c}}{||\mathbf{c}||}$. Thus you can write $\mathbf{x}$ as

$$\mathbf{x} = \frac{s}{||\mathbf{c}||} \hat{\mathbf{c}} + \mathbf{w}$$

where $\mathbf{w}$ is any vector orthogonal to $\hat{\mathbf{c}}$. That is, the locus of such $\mathbf{x}$ is a plane with normal $\hat{\mathbf{c}}$.

Let $c_1$ be the unit vector along $c$ direction, let $I$ be the unit matrix. Then the vector $w$ in the previous answer can be expressed as $$w=b.(I-c_1c_1)$$. Where $b$ is an arbitrary vector.

Thus $w.c=0$.

Sorry, I typed everything from my iPhone, which is not convenient.