Write down the cartesian equation of the plane that contains the point (1,0,3) and the vectors $$u=(-1,1,2)$$ $$v=(1,0,-1)$$ So I cross product-ed vectors $v$ and $u$, substitute the point into the normal vector, and came up with this equation $$-x+y-z=-4$$ I'm not sure whether I'm doing this correctly so I need help with checking, thank you guys!
Edited: And would the vector equation of the plane be $$r=(1,0,3)+s(-1,1,2)+t(1,0,-1)$$
We can find the normal vector $\vec{\mathbf n}$ to the plane by finding the cross product of the vectors $\vec{\mathbf u}$ and $\vec{\mathbf v}$. $$\vec{\mathbf u}\times\vec{\mathbf v}=\langle(1)(-1)-(2)(0),(2)(1)-(-1)(-1),(-1)(0)-(1)(1)\rangle=\langle-1,1,-1\rangle$$ This comes from the fact that $\vec{\mathbf u}\times\vec{\mathbf v}=\langle u_2v_3-u_3v_2,u_3v_1-u_1v_3,u_1v_2-u_2v_1\rangle$ [1]. We can place the components of this normal vector into the general equation of a plane [2]. $$-x+y-z+d=0$$ We can easily solve for $d$ by placing the coordinates of point $P(1,0,3)$ into the equation. $$-(1)+(0)-(3)+d=0\implies-4+d=0\implies d=4$$ We now have the general equation of the plane containing the two vectors $\vec{\mathbf u}$ and $\vec{\mathbf v}$ and the point $P(1,0,3)$. $$-x+y-z+4=0$$ We have the normal vector $\vec{\mathbf n}=\langle-1,1,-1\rangle$ to the plane and the position vector $\vec{\mathbf r}_0=\langle1,0,3\rangle$ of the point $P(1,0,3)$ in the plane, so we can find the vector equation of the plane [3]. $$\vec{\mathbf n}\cdot\left(\vec{\mathbf r}-\vec{\mathbf r}_0\right)=0\implies\langle-1,1,-1\rangle\cdot\left(\vec{\mathbf r}-\langle1,0,3\rangle\right)=0$$ Here, $\vec{\mathbf r}$ is the position vector of some arbitrary point in the plane. For example, the point $Q(2,-1,1)$ is in the plane because $-(2)+(-1)-(1)+4=0$. $$\langle-1,1,-1\rangle\cdot\langle(2-1),(-1-0),(1-3)\rangle=(-1)(1)+(1)(-1)+(-1)(-2)=0$$