Find an equation for the plane containing $u$ and $v$. Consider the vectors $u = (3, 1, 0)$ and $v = (3, 0, 1)$
Find an equation for the plane containing $u$ and $v$:
Can I assume that the $(x,y,z) = (0,0,0)$ because there is no point specified.
Cross Product gives $(1,-3,-3)$ so:
$$ax + by + cz = 0$$
$$(1,-3,-3)\cdot((x-0),(y-0),(z-0))= x -3y -3z = 0$$
Please comment...
 A: Given two linearly independent vectors $\mathbf{u},\mathbf{v}\in\mathbb{R^3}$, then the cross product $\mathbf{u\times v}$ gives a vector which is perpendicular to both vectors, thus forming a normal to the plane passing through the origin.
The dot product can be used to test for perpendicularity, $\mathbf{a\cdot b}=0$ iff $\mathbf{a}\perp \mathbf{b}$. This can be seen in that $\mathbf{(u\times v)\cdot u}=\mathbf{(u\times v)\cdot v}=0$.
This leads to the equation of a plane spanned by $\mathbf{u},\mathbf{v}$ and passing through the origin as $\mathbf{(u\times v)\cdot x}$, where $\mathbf{x}=(x,y,z)$.
A: Let $(x,y,z)$ be in the same plane as $u$ and $v$. Note that $u$ and $v$ are linearly independent, so they will form a basis in the plane spun by themselves. Since three vectors in a plane cannot be linearly independent, it follows that $(x,y,z), u, v$ must be linearly dependent, which you can write using the determinant as
$$\left| \begin{array} {ccc}
x & y & z \\
3 & 1 & 0 \\
3 & 0 & 1
\end{array} \right| = 0 ,$$
which can be rewritten as $x-3y-3z = 0$, and this is the equation that you are looking for.
