Find the vector equation for a plane with cartesian equation $x-3y+2z=5$.
My work.
I have managed to use vector techniques to find the equation as follows:
$x=5+3y-2z$
$$(x,y,z)=(5+3y-2z, y, z)$$
Factoring out the $y$ and $z$,
$$(x,y,z)=(5,0,0)+y(3,1,0)+z(-2,0,1).$$
In terms of scalars:
$$(x,y,z)=(5,0,0)+s(3,1,0)+t(-2,0,1).$$
However, our lecturer emphasized on trying to use linear algebra, particularly Gauss reduction, to find the vector equation of planes from cartesian. I'm not too sure of how to go around this. Could someone please help me use linear algebra for this question? Thanks:)
Best Answer
let us compute three points of this equation $$x-3y+2z=5$$ for example $$P_1(5,0,0),P_2(0,1,4),P_3(1,0,2)$$ then you can compute the equation of the plane as follows: $$[x,y,z]=(5,0,0)+\alpha(-5,1,4)+\beta(-4,0,2)$$ with real numbers $$\alpha,\beta$$