Let $v,w$ and $v'$ be nonzero vectors such that $\langle v,v'\rangle=0$ and $\langle w,v'\rangle =0$ then $\langle v\rangle = \langle w\rangle$

Question

Let $v,w,v'$ be nonzero vectors in $\Bbb R^2$ such that $v\cdot v'=0$ and $w\cdot v'=0$ then $\Vert v\Vert=\Vert w\Vert$.

how to prove this :

where to start $\langle v', v+w\rangle=\langle v',w\rangle +\langle v,v\rangle =0+0=0$.

Answer 1

It’s not true

EG $v=(1,0)$, $w=(2,0)$, $v’=(0,1)$.

What is true is that $v=\lambda w$ $\lambda \in \mathbb{R}$

Answer 2

The notation is quite confusing, but here I think $\langle v\rangle$ denotes the vector space spanned by $v$, while $\langle v,v'\rangle$ is the scalar product of $v$ and $v'$.

So the question really becomes : if two non-zero vectors of $\Bbb R^2$ are orthogonal to the same non-zero vector, then they must be colinear.

To prove it, it suffices to note that $\langle v,v'\rangle =0=\langle w,v'\rangle $ implies that $v,w\in \operatorname{span}(\{v'\})^\bot$, or equivalently that $$\operatorname{span}(\{v\})\subseteq\operatorname{span}(\{v'\})^\bot\supseteq\operatorname{span}(\{w\}),$$ and use the fact that all these spaces are one-dimensional to conclude.