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 :)

  • $\begingroup$ 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 $b$ $\endgroup$ – J. W. Tanner Apr 11 at 4:34
  • $\begingroup$ 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? $\endgroup$ – qwerty2019 Apr 11 at 4:41
  • $\begingroup$ $(l,m,n)$ 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? $\endgroup$ – J. W. Tanner Apr 11 at 4:43
  • $\begingroup$ 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? $\endgroup$ – qwerty2019 Apr 11 at 4:47
  • $\begingroup$ Your solution is correct, but I'm not sure I understand your question -- the solutions are {$(0,c,-c)|c\in\mathbb R$} $\endgroup$ – 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.

  • $\begingroup$ Wow thank you so much for your patience. I finally understand what you mean. Hope you have a great day! $\endgroup$ – qwerty2019 Apr 11 at 4:57

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 $$ \{ (0,-c,c) \in \mathbb{R}^3 \mid c \in \mathbb{R} \}.$$

  • $\begingroup$ Thanks for your help! $\endgroup$ – qwerty2019 Apr 14 at 14:32

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