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 ?

  • 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

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?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.