# How to write vectors in abbreviated set notation?

I was wondering whether anyone knew how to write a vectors in abbreviated set notation to express the solutions to this question:

"Determine all values of x, y, z ∈ R such that (x, y, z) is perpendicular to both a = (1, 1, 1) and b = (−1, 1, 1)."

Letting n=[x , y, z], I figured out the two simultaneous equations (we have not covered cross product yet)

1. x + y + z = 0
2. -x + y + z = 0

However, the question wants us to express the answer in the form of {...|c ∈ R) which I am unsure how to do.

I understanding expressing the answer in the regular notation would be something like {(x, y , z) ∈ $$R^3$$ | x=0 and y=-z}.

Thank you very much for your help guys! Much appreciated :)

• If you find a vector $(l,m,n)$ perpendicular to $a$ and $b$, then {$c(l,m,n)|c\in \mathbb R$}$=${$(cl,cm,cn)|c\in\mathbb R$} is a set of vectors perpendicular to $a$ and $b$ – J. W. Tanner Apr 11 at 4:34
• Hi, thanks for your help! I understand your explanation for a given vector; however, in this case, I have infinite solutions, so how would I express that in the abbreviated set notation? – qwerty2019 Apr 11 at 4:41
• $(l,m,n)$ is a given vector; {$c(l,m,n)|c\in\mathbb R$} is an infinite collection of vectors, since $\mathbb R$ is infinite; what are the solutions you have? – J. W. Tanner Apr 11 at 4:43
• Thank you for your quick response. In this case, I think my solution for vectors which are perpendicular to a and b must satisfy l=0 and m= -n. However, if I write, {c(l,m,n)|c∈R}, wouldn't that mean l is allowed to equal 0? – qwerty2019 Apr 11 at 4:47
• Your solution is correct, but I'm not sure I understand your question -- the solutions are {$(0,c,-c)|c\in\mathbb R$} – J. W. Tanner Apr 11 at 4:51

There are infinitely many vectors $$(x,y,z)$$ perpendicular to $$(1,1,1)$$ and $$(-1,1,1)$$;
if you find one of them, say ($$l,m,n$$), then all vectors in the set {$$(cl,cm,cn)|c\in \mathbb R$$} are solutions.
You found a solution $$(0,1,-1),$$ so {$$(0,c,-c)|c\in\mathbb R$$} are solutions.
The reduced (row) echelon form of the homogeneous linear system above (equations 1. and 2.) is $$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \end{bmatrix},$$ which tells us that $$x = 0$$ and $$y = -z$$. By convention, $$z$$ is a free variable, so every solution is of the form $$\{ (0,-c,c) \in \mathbb{R}^3 \mid c \in \mathbb{R} \}.$$