# Visualise all vectors perpendicular to one vector and two vectors in R^3 [Strang P19 1.2.6]

I'm only asking about visual/geometric solutions herein.
(b) The vectors perpendicular to any vector in $\mathbb{R^3}$ lie on what?.
(c) The vectors perpendicular to any two vectors in $\mathbb{R^3}$ lie on what?.

(b) The algebra proves: $\forall \, (a, b, c) \in \mathbb{R^3}$, $(a,b,c) \cdot (x, y, z) = 0 \implies ax + by + cz = 0$, a plane through the origin with normal vector $(a, b, c)$. How do I see this from (descrying) this? I sketched the blue, green, orange, and red vectors to be parallel to the position vector $p$. Yet I can't see how these four vectors lie on the same plane?

(c) I can see that this is true pictorially, thanks to the cross product. I only need to extend the cross product vector infinitely at both ends to form the line $\perp$ to any two vectors in $\mathbb{R^3}$.

## 2 Answers

(b) It's $\Bbb R^3$, our dimension. Consider yourself as a vector, then you are perpendicular on what? Of course the flat land that you stand on it, a plane. (Don't forget if the vector is zero then the answer is the whole $\Bbb R^3$.)

They don't need to line on the same plane, in fact:

1. Take out a pencil, and an A4 paper. Now, imagine that your pencil is a vector, the problem is to find where to put your A4 paper, so that the 2 are perpendicular. There are infinitely many solutions to this problem, right? And all the possible positions of your A4 (that are solutions to the above problem) form many layers parallel to each other. And of course, if you put another pencil on the A4 paper, it should be perpendicular to the former pencil.
2. Of course, one such layer passes thru the origin.
3. Don't forget that you can move your vector parallelly form one layer, to any other layer.

Regards,