# Prove true or false: $\vec{u}\cdot \vec{v} = \vec{u} \cdot \vec{w} \implies \vec{v} = \vec{w}$

## Problem

Prove true or false following statement:

$$\vec{u}\cdot \vec{v} = \vec{u} \cdot \vec{w} \implies \vec{v} = \vec{w}$$ when $$\{\vec{u},\vec{v},\vec{w}\} \in \mathbb{R}^n, n \in \mathbb{N}$$

## Attempt to solve

If I take dot product with $$\vec{u}$$ from both sides i get:

$$\vec{u}\cdot(\vec{u}\cdot\vec{v})=\vec{u}(\vec{u}\cdot\vec{w}) \implies ||\vec{u}||^2 \cdot \vec{v} = ||\vec{u}||^2 \cdot \vec{w} \implies \vec{v} = \vec{w}$$

However not quite sure if this part is correct

$$\vec{u}\cdot(\vec{u}\cdot\vec{v})=\vec{u}(\vec{u}\cdot\vec{w}) \implies ||\vec{u}||^2 \cdot \vec{v} = ||\vec{u}||^2 \cdot \vec{w}$$

How do you define dot product of $$\vec{u}\cdot \vec{u} \cdot \vec{v}$$

$$\begin{bmatrix} a \\b\end{bmatrix} \cdot \begin{bmatrix} a \\b\end{bmatrix} \cdot \begin{bmatrix} c \\ d \end{bmatrix}= a^2\cdot c+ b^2 \cdot d$$

$$\begin{bmatrix} a \\b\end{bmatrix} \cdot (\begin{bmatrix} a \\b\end{bmatrix} \cdot \begin{bmatrix} c \\ d \end{bmatrix} )= (ac+bd) \begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} a(ac+bd) \\ b(ac+bd) \end{bmatrix}$$

Which is quite confusing. I know how dot product is defined with two vectors

$$\begin{bmatrix} a \\ b \end{bmatrix} \cdot \begin{bmatrix} c \\ b \end{bmatrix} = ac+bd$$

But i don't quite get how it would work for 3 vectors. Another way to solve this would be try to divide with vector $$\vec{u}$$ which is same as multiplying $$\vec{u}$$ with its inverse vector ? But I have no clue how do you define inverse vector ?

• There's no definition of $\vec{u}\cdot \vec{u}\cdot \vec{v}$ but for $(\vec{u}\cdot \vec{u})\cdot \vec{v}$ or $\vec{u}\cdot (\vec{u}\cdot \vec{v})$ there is one. – Fakemistake Oct 27 '18 at 12:37
• On the other hand, if $\vec{v}$ and $\vec{w}$ are vectors in $\Bbb{R}^n$, the following are equivalent: (1) $\vec{u}\cdot \vec{v}=\vec{u}\cdot\vec{w}$ for every $\vec{u}\in\Bbb{R}^n$; (2) $\vec{u}\cdot \vec{v}=\vec{u}\cdot\vec{w}$ for $n$ linearly independent $\vec{u}\in\Bbb{R}^n$, then $\vec{v}=\vec{w}$; (3) $\vec{v}=\vec{w}$. – user593746 Oct 27 '18 at 16:26
• This is trivially false for $\vec u=0$. – amd Oct 27 '18 at 20:05

False. Take $$u=(1,0,0), v=(0,1,0), w=(0,0,1)$$.

Consider two distinct vectors $$\vec{v} \neq \vec{w}$$ in $$\mathbb{R}^3$$. They are both perpendicular to the vector $$\vec{u} := \vec{v} \times \vec{w}$$, so $$\vec{u} \cdot \vec{v} = \vec{u} \cdot \vec{w} = 0$$, but $$\vec{v} \neq \vec{w}$$, so the statement is false.

In fact, it is easy to see that the scalar product distributes under addition. In particular, for $$\vec{u}$$, $$\vec{v}$$, $$\vec{w}$$ vectors in $$\mathbb{R}^n$$, $$\vec{u} \cdot \vec{v} = \vec{u} \cdot \vec{w} \iff \vec{u} \cdot \left(\vec{v} - \vec{w}\right) = 0,$$ so the equality holds iff the vector $$\vec{v} - \vec{w}$$ is either zero (in which case $$\vec{v} = \vec{w}$$) or it is orthogonal to $$\vec{u}$$. You can construct a load of vectors orthogonal to a given vector in $$\mathbb{R}^n$$, so you can construct a load of counterexamples.

$$\vec u \cdot (\vec v -\vec w)= 0.$$

$$\vec a := \vec v -\vec w$$ is perpendicular to $$\vec u$$.

In $$2D$$: $$\vec u =(1,0)$$; $$\vec a = (0,1)$$.

Then $$\vec v =(0,1)+ \vec w.$$

Are they equal?