Find the basis in the plane through the origin with normal vector <1, 2, 3>
My book is very vague when it comes to explaining methods of finding a basis, let alone using a normal vector. I found this example and would like some pointers on how to solve a problem such as this one.
Find a basis for $\mathbb{R}^3$ with one vector being $[1,2,3]^\intercal$.
Apply Gram-Schmidt process to the above basis. Make sure to use $[1,2,3]^\intercal$ as the first vector.
Remove $[1,2,3]^\intercal$ from the resulting basis.
Henry W's answer is good, but time consuming. If you are only working in $\mathbb R^3$, you can choose an arbitrary vector $v\in\mathbb R^3$ which is not a multiple of $\langle 1,2,3\rangle$. Let $a=\langle 1,2,3\rangle\times v$ and $b=\langle 1,2,3\rangle\times a$. The vectors $a$ and $b$ are a basis for the plane normal to $\langle 1,2,3\rangle$.
The plane is the solution set of $x_1+2x_2+3x_3=0$, i.e., the kernel of the linear map $T:(x,y,z)^T\mapsto x_1+2x_2+3x_3$. A basis for the kernel of a linear map can be read directly from the row-reduced echelon form of its matrix (you can find a description of how to do this here). In this case, the matrix of $T$ is just the given row vector. It’s already rref, so a basis for the kernel can be found immediately: $\{(2,-1,0)^T,(3,0,-1)^T\}$.
