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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 ?

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  • 1
    $\begingroup$ 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. $\endgroup$ – Fakemistake Oct 27 '18 at 12:37
  • $\begingroup$ 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) $\vec{v}=\vec{w}$. $\endgroup$ – user593746 Oct 27 '18 at 16:26
  • $\begingroup$ This is trivially false for $\vec u=0$. $\endgroup$ – amd Oct 27 '18 at 20:05
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False. Take $u=(1,0,0), v=(0,1,0), w=(0,0,1)$.

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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.

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

$\vec a := \vec v -\vec w$ 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?

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