# Find two vectors orthogonal to $v = [1,2,0]$

Find two vectors orthogonal to $v = [1,2,0]$

Usually i see questions with asking you two find given two vectors find two orthogonal vectors for it. Then you would use cross product and then use the result to find the unit vector.

What i do not understand is how would i do this for a single vector if i'm trying to find two vectors orthogonal to it

• I would just do it in my head to by honest... $u=[2,-1,0]$ and $-u$ come to mind straight away. There are a vast number of course --- they form a plane. Nov 18 '15 at 17:12
• A really cheap one is $(0,0,1)$. Nov 18 '15 at 17:14
• @AndréNicolas Oh i see so it could literally be anything perpendicular to it. So there wouldnt be a need to use a fancy formula? Nov 18 '15 at 17:17
• Take your pick. Any $x$, $y$, $z$ that satisfies $x+2y=0$ will do. What if I asked you to find two numbers whose sum is $10$? Nov 18 '15 at 17:17
• More interesting is to find two non-zero vectors perpendicular to your given one and also to each other. For that, playing around will also do it, but for larger similar problems you do need some machinery. Nov 18 '15 at 17:21

Denote "$v$ is orthogonal to $w$" by $v\ \bot\ w$. Then we define orthogonality by $$v\ \bot\ w \iff v\cdot w=0$$ where $v\cdot w$ is the dot product of $v,w \in \Bbb R^n$.

So a vector $(x,y,z)$ is orthogonal to $v$ if $$(x,y,z) \cdot (1,2,0)=x+2y=0$$

Clearly there are no restrictions on $z$ so you can pick any value of $z$. But $x$ and $y$ are related by the equation $x=-2y$.

So just pick any values of $x$'s (or $y$'s) and $z$'s and use that equation to find the last coordinate of suitable vectors.

For instance $(2,-1,0)$, $(-10,5,3)$, $(2\pi,-\pi, e)$, $(0,0,\frac {11}{3})$, etc are all vectors orthogonal to $v$. Confirm this for yourself.

"Orthogonal" also means perpendicular so all you would need is two vectors perpendicular to $v = [1,2,0]$, so the answers could be -v which is $v = [-1, -2, 0]$ and $[0,0,1]$

• $-v$ is not orthogonal to $v$.
– user137731
Nov 18 '15 at 17:55