I am given vector $a=(2,-1,3)$ and vector $b=(3,0,-1)$. I need to find the equation of a plane which passes through this 2 vectors.
I tried finding the cross product and I get vector $(1, 11, 3)$. If this vector is normal to the plane then $$x+11y+3z+D=0$$ is the equation of the plane.
But how do we find $D$?
On
$\textbf{Hint: }$ $$$$I assume you are to find the plane containing the $lines$ parallel to the vectors $\vec{a}=2i-j+3k$ and $\vec b=3i-k$. Without this assumption, the question cannot be solved beyond what you have already reached. $$$$ Let $\vec{r}$ be the position vector of any point in the plane. let $\vec{p}$ be the position vector of the point of intersection of the two (non parallel) lines that have been given. $$$$ Clearly $\vec{r}-\vec{p}$ lies in the plane, hence it is perpendicular to the normal to the plane (given by the cross product of $2i-j+3k$ and $3i-k$) . Hence, $$(\vec{r}-\vec{p}).\left((2i-j+3k) \times (3i-k)\right)=0$$
Yes you derivation by cross product is correct.
Without any other information, we assume that the plane is throught the origin thus $d=0$.
Otherwise, for the general case, we need an extra information on a point $P(x_0,y_0,z_0)\in$ plane or by the distance from the origin in order to find $d$.