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?
 A: $\vec x\perp\vec z$: $x_1z_1+x_2z_2=0$
$\vec y\perp\vec z$: $y_1z_1+y_2z_2=0$
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.
A: 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 $x$.
A: 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 $W$
A: $\mathbb{R}^2$ 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:
$${\vec y}=\sum_{i=1}^2({\vec y}\cdot{\vec e}_i){\vec e}_i=({\vec y}\cdot{\vec x}){\vec x}+({\vec y}\cdot{\vec z}){\vec z}=({\vec y}\cdot{\vec x}){\vec x}=\alpha{\vec x}$$
A: 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 $t$.)
