# Proof: $\vec x,\vec y \perp \vec z \Rightarrow \vec x || \vec y$

I have to prove $\vec x \perp \vec z$ and $\vec y \perp \vec z$ imply $\vec x || \vec y$ where $\vec x,\vec y,\vec z \in \mathbb{R}^2$ and $z$ nonzero.

I know $x \perp z \Leftrightarrow x_1z_1+x_2z_2=0$ and $y \perp z \Leftrightarrow y_1z_1+y_2z_2=0$. If two vectors are parallel, I can write $\vec x = \alpha \vec y$.

I tried to write $x_1z_1+x_2z_2=y_1z_1+y_2z_2$ but this didn't help me to find an $\alpha$ to satisfy $\vec x = \alpha \vec y$.

Anyone got a hint for me?

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What is the vector space in the question ? – Belgi Apr 29 '13 at 10:10
I assume you're working in two dimensions here. For a three-dimensional space, your statement clearly doesn't hold, since $(1,0,0)$, $(0,1,0)$, $(0,0,01)$ are all perdendicular to each other... You might want to mention that in the question... – fgp Apr 29 '13 at 10:11
It's for $\mathbb{R}^2$, sorry – ulead86 Apr 29 '13 at 10:11
I added the condition that $z\neq 0$ since the statement is clearly false if $z=0$. – rschwieb Apr 29 '13 at 10:44
$z1,z2 \not=0 \to y_1x_1z_1+y_1x_2z_2-x_1y_1z_1-x_1y_2z_2=0 \to y_1x_2-x_1y_2=0 \to \dfrac{x_1}{y_1}=\dfrac{x_2}{y_2}=\alpha \to \vec x = \alpha \vec y$ – chenbai May 3 '13 at 3:31

## 5 Answers

If $x=0$ then $y$ is degenerately parallel to $x$ in the sense that $0y=x$. We suppose now that $x\neq 0$.

Since $x$ and $z$ are nonzero and perpendicular, they form a basis for $\Bbb R^2$. That means it is possible to write $y=\alpha x+\beta z$ for some $\alpha, \beta\in \Bbb R$.

Now we check the inner product of $y$ with $z$ and use the given facts that $y\cdot z=0$ and $x\cdot z=0$:

$$0=y\cdot z=\alpha (x\cdot z) +\beta (z\cdot z)=0+\beta \|z\|^2$$

Since $z$ is nonzero $0=\beta\|z\|^2$ implies $\beta=0$. Thus in the original expression $y=\alpha x+0$ yields that $y$ is a scalar multiple of $x$.

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(+1) for a very simple proof. I enjoy reading proofs like this. Thanks. – BU982T Apr 29 '13 at 10:47

$\vec x\perp\vec z$: $x_1z_1+x_2z_2=0$

$\vec y\perp\vec z$: $y_1z_1+y_2z_2=0$

With matrix notation: $$\left( \begin{array}{cc} x_1&x_2\\ y_1&y_2 \end{array}\right)\cdot\left(\begin{array}{c} z_1\\ z_2 \end{array}\right)=\left(\begin{array}{c} 0\\ 0 \end{array}\right)$$

Suppose $\vec x,\vec y$ not parallel. Then the matrix is invertible, and this implies (multiply both sides by the inverse matrix) that $\vec z=0$, which I suppose is not the case.

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Hint: If $W$ is a vector subspace of a vector space $V$ then $$\dim(W^{\perp})=\dim(V)-\dim(W)$$ where $W^{\perp}$ is the set of all vectors in $V$ that are perpendicular to all the vectors in $W$

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good idea, but is it possible to find a proof without vector subspaces? – ulead86 Apr 29 '13 at 10:15
@ulead86 It is. – rschwieb Apr 29 '13 at 10:31

$\mathbb{R}^2$ is a flat two dimensional space. Therefore, any two vectors orthogonal to each other span an orthogonal basis in $\mathbb{R}^2$. Now choose, say, ${\vec e}_1 ={\vec x}$ and ${\vec e}_2 ={\vec z}$ as basis vectors for your $\mathbb{R}^2$. A basis of a flat space has the property that any vector in the space can be decomposed into components along the basis vectors (through a projection via scalar product). So that you can write:

$${\vec y}=\sum_{i=1}^2({\vec y}\cdot{\vec e}_i){\vec e}_i=({\vec y}\cdot{\vec x}){\vec x}+({\vec y}\cdot{\vec z}){\vec z}=({\vec y}\cdot{\vec x}){\vec x}=\alpha{\vec x}$$

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Solve your system \begin{align} x_1 z_1 + x_2 z_2 &= 0 \\ y_1 z_1 + y_2 z_2 &= 0 \end{align} for $z_1$ and $z_2$, but note that $z_1$ and $z_2$ aren't both zero. You'll see, then, that there must be a relation between the $x_i$ and $y_i$; you can tease this relation into the form $x = \alpha y$. (Hint: If $pq-rs=0$, then $pq=rs=t$ for some $t$.)

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