MHB Proving Parallelism of Vectors with Perpendicularity Constraints

[Ganesh Ujwal]

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 $\vec x = \alpha \vec y$.

 

[Opalg]

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 $\vec x = \alpha \vec y$.

If $\vec x = 0$ the result is true, because then $\vec x = 0.\vec y$. So suppose that $\vec x \ne0$. Then $\vec x$ and $\vec z$ form a basis for $\mathbb{R}^2$ (because they are two linearly independent vectors in a two-dimensional space). Therefore $\vec y$ must be a linear combination of $\vec x$ and $\vec z$. Can you take it from there?