Given two vectors $u,v\in\mathbb{R}^3$, find the parametric equations for the plane through the origin parallel to $u$ and $v$.
I'm having a difficult time understanding this question. How can a plane through the origin in $\mathbb{R}^3$ be parallel to two vectors, both of which start from the origin? Surely they all intersect at the origin and thus not parallel?
More concretely, $u=(1,2,3)$ and $v=(-2,1,1)$. The cartesian equation of the plane is $x+7y-5z=0$, which is essentially the solution. But I want to know why.
Best Answer
Plane:
Generally: $ \vec r = \vec a + t \vec u + s \vec v$;
This plane passes through the point
specified by $\vec a$.
The direction vectors $ \vec u $ and $\vec v$,
starting from this point, vector addition, span the plane.
$\vec a = (0,0,0), \vec u = (1,2,3)$ , and
$\vec v= (-2,1,1).$
For the problem (solution)
the normal vector $ \vec n $, perpendicular
to both direction vectors is required.
Cross product of vectors:
$\vec n = \vec u × \vec v = (-1,-7,5)$;
Plane:
$\vec n \cdot ( \vec r - \vec a) = 0$,
$(-1,-7,5) \cdot (x,y,z) =0$.
Finally: $ -x -7y +5z = 0.$
Helps?