Find a pair of basis vectors for the plane given by the equation x+2y+3z=0
I was given the Hint: you can find points on the plane by choosing values for two variables and solving for the third
I am in calc III and doing this on the program Mathematica
Any two vectors lying on the plane will do as long as they're linearly independent. A simple way to find such vectors is, as mentioned to choose two variable and solve for the third:
let $x=y=1$, then by substitution into the equation of the plane, $z=-1$
This tells you the vector $(1,1,-1)$ lies on the given plane. You can alsocheck your answer by substituting the values back in and making sure the equation is consistent: $$(1)+2(1)+3(-1)=0$$ Then all we need to do is find another vector which satisfies this equation and isn't linearly dependent on the one we've already found.
let $x=z=1$, then by substitution, $y=-2$, so the vector $(1,-2,1)$ should lie on the plane. We'll check our answer again: $$(1)+2(-2)+3(1)=0$$ Again, this is consistent, so we must have found a valid vector. So, we have $(1,1,-1)$ and $(1,-2,1)$, how do we know they're linearly independent? the only way just two vectors can be linearly dependent is if they're scalar multiples of eachother, which these two vectors are obviously not. So, a valid basis for $x+2y+3z=0$ is given by: $$B=\{(1,1,-1),(1,-2,1)\}$$
Note however, that any two linearly independent vectors lying on the given plane form a valid basis for it, its just convenient to find two such vectors via this method.