Find a basis for the orthogonal complement of the plane $2x + 3y + 4z = 0$.

I am very confused on how to solve this problem, I know what orthogonal complement and a basis are but I don't understand how to represent the plane as a matrix.
I know there is an easy way of doing this with the fundamental theorem of linear algebra but I want to learn how to do it the long way and how to think through the problem. How do I get started?



The orthogonal complement of a subspace $U$ is the collection of all vectors $v$ such that $v\cdot u=0$ for all $u\in U$. Let $v$ be a vector in the orthogonal complement, given by $(a,b,c)$. Let $u$ be a vector in the plane given by $(x,y,z)$. Then $v\cdot u=0$, or $ax+by+cz=0$. What values could $a$, $b$, and $c$ have?

  • $\begingroup$ $[2, 3, 4]$, the orthogonal complement is similar to saying the nullspace of the plane, so the $x$, $y$, and $z$ values that make that equation $0$ are vectors in the plane? $\endgroup$ – idknuttin Apr 3 '16 at 2:17
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    $\begingroup$ Yes, and the orthogonal complement is the span of $(2,3,4) $. $\endgroup$ – Alex S Apr 3 '16 at 2:32

Your plane is a two dimensional subspace in $\mathbb{R}^3$, so the orthogonal compliment is a one dimensional object in $\mathbb{R}^3$. We are looking for a subspace such that each vector in the plane is orthogonal to every vector in the orthogonal compliment.

Do you know of any structure associated with planes that is orthogonal to vectors in a plane?

Scroll over for the answer:

The Normal Vector


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