If a plane passes through two points:
$A=(1,0,-2)$ and $B=(3,-1,1)$
and is parallel to the line:
$r=(3,-1,5)^T + a(0,2,-1)^T$,
then what is the vector equation of the plane?
I have obtained that the plane must be of the form:
$p=(1,0,-2)^T +\delta_1((1,0,-2)^T-(3,-1,1)^T)+\delta_2((3,-1,1)^T-(x,y,z)^T)$
but I can't determine what direction vectors of the plane are parallel to the direction vector of the line?
e.g
is it true to say $(3,-1,1)^T-(x,y,z)^T=\mu(0,2,-1)$?
Best Answer
the plane contain the line $AB$ and the vector director of the line $r$ which is $\vec{u}=(0,2,-1)$.
the cross product $ \vec{u}\times \vec{AB}$ gives the normal vector to the plane.
which is
$(-5,2,4)$
so the cartesian equation of the plane has the form
$-5x+2y+4z=D$.
the point $A(1,0,-2)$ is in the plane, so $-5-8=-13=D$.