How do I find the orthogonal basis for this plane? Question: 
$P$ is a plane through the origin given by 
$x + y + 2z = 0$.
Find an orthogonal basis v1, v2  ∈ $P$.
My answer:
I'm assuming the question asks for two vectors that span this plane $P$. But the chapter that this problem is for doesn't say anything about the $x,y,z$ equation of a plane that was given here...so I did some searching online and learned that this helps find the "normal vector".
In this case it would be $n = (1,1,2)$, right?
Then if all the vectors that span this plane are orthogonal to the normal vector, I can use the dot product.
I chose the following two vectors:
v1 = $(1,1,-1)$
v2 = $(3,3,-3)$
Was this question answered correctly?
 A: First, find a basis for the plane as
$$ x + y + 2z = 0 \implies x=-y-2z \implies (x,y,z)=(-y-2z,y,z)$$
$$ \implies (-y,y,0)+(-2z,0,z)\implies y(-1,1,0)+z(-2,0,1). $$
So, our basis is $\left\{(-1,1,0),(-2,1,0)\right\}$. Now, you need to apply Gram-Schmidt process to the basis set to get the orthogonal basis.
A: Your solution is not correct as $v_1$ and $v_2$ are linearly dependent. However, your approach is correct. We need to find just another vector orthogonal to $n$ which is not a scalar multiple of $v_1$. Note that $(0,1,-2)$ is such a vector.
A: The question asks for two vectors in the plane $P$ that  are orthogonal to each other. You can choose any vector $\vec{v_1}$ as the first vector, but not any vector $\vec{v_2}$. In fact, there are two vectors that are in the plane $P$ and that are orthogonal to $\vec{v_1}$ (try to find out why).
Furthermore, any vector in $P$ will be orthogonal to $\vec{n}$, by definition. We are thus looking for a vector $\vec{v_2}$ that is orthogonal to both $\vec{v_1}$ and $\vec{n}$.
One of the ways to find $\vec{v_2}$ is to use the vectorial product...
A: Chose $v_1=(-1,1,0)$ and $v_2=v_1\times n$.  No Gram-Schmidt needed in this case.
A: $n = (1, 1, 2)$ - normal to plane
$v_1 = (1,1,-1)$ - first base vector on plane
$v_2 = n \times v_1 =  (-3,3,0)$ - second base vector on plane
