Do there exist three non null vectors a,b,c with $a.b=a.c$ such that $b\ne c$?

Question

Do there exist three non null vectors $a,b,c$ with $$a.b=a.c$$ such that $b\ne c$?

My attempt:

a.(b-c)=0

But is it possible to say b-c=0 since a is non null?

Answer 1

Consider $$a=\begin{bmatrix}1\\0\end{bmatrix}, b=\begin{bmatrix}0\\1\end{bmatrix},c=\begin{bmatrix}0\\2\end{bmatrix}.$$ As long as $a$ is perpendicular to $b-c$, you will get the equality $a\cdot b=a\cdot c$.

Answer 2

$\vec a\cdot (\vec b-\vec c)=0.$

$\vec a$ is perpendicular to $(\vec b-\vec c)$.

In 2D:

Let $\vec a =(1,0)$, then chose $\vec b-\vec c =(0,1)$(why?).

$\vec a=(1,0)$; $\vec b=\vec c +(0,1)$;

Answer 3

Take isosceles triangle $ABC$ such that $AB = AC$ and let $M$ be a midpoint of $BC$. Then $$\vec{AM} \cdot \vec{AB} = \vec{AM} \cdot \vec{AC}$$