Parametric form of a plane given a point and perpendicular vector 
My apologies if a question like this already exists but I haven't been able to find anything. A point is given (1,4,5) and a perpendicular vector is also given (7,1,4). I am asked to find the parametric (vector) form of a plane using this information.


I understand that the parametric form is given by $\ \vec r=\vec r_0+s \vec u +t \vec v; \ s,t \ \epsilon \ \Bbb R$ and that in this case $\vec r_0$ will be (1,4,5). I was able to find the Cartesian equation to be $7x+y+4z=31$ and the answer for the parametric form is given as $\ \vec r=(1,4,5)+s(4,0,-7)+t(0,4,-1)$
I appreciate any help that anyone could give. Thanks
 A: $\vec u(x,y,z)$ and $\vec v(x',y',z')$ are non-colinear vectors parallel to the plane, i.e. $\vec n\cdot\vec u=\vec n\cdot\vec v=0$, where $\vec n=(7,1,4)$ is a vector normal to the plane. We have$$\vec n\cdot\vec u=7x+y+4z=0$$and similarly for $\vec v$. You need to find suitable values of $x,y,z$ that satisfy the above.
A: In $\ \vec r=\vec r_0+s \vec u +t \vec v$, $\vec r_0$ is the position vector of a point on the plane, $\vec u$ and $\vec v$ are two vectors parallel to the plane. There are infinitely many choices for each one of them. $\vec r_0=(1,4,5)$ is a convenient choice. Both $\vec u$ and $\vec v$ should be perpendicular to $(7,1,4)$, and they should be linear independent. $\vec u=(4,0,-7)$ and $\vec v=(0,4,-1)$ are convenient choices as $(7,1,4)\cdot(4,0,-7)=0$ and $(7,1,4)\cdot(0,4,-1)=0$.
$(1,-7,0)$ is also an easy choice. Actually, we can take arbitrarily two coordinates and decide the third coordinate by the dot product. For example, if we take $x=1$ and $z=-2$, then $(7,1,4)\cdot(x,y,z)=0$ will yield $y=1$. $(1,1,-2)$ can also be a candidate for $\vec u$ or $\vec v$.
