Find the equation of the plane in $\mathbb{R}^3$ that contains the points $v=(1, 2, 1)$, $w=(0, 1, 4)$, and
$q=(-1, -1, 0)$.
Equation of the plane is given by $ax+by+cz+d=0$ with normal vector $N=(a,b,c)$.
Let $t=w-v=(2,3,1)$ and $s=q-v=(1,2,4)$.
Then I calculate $q=N-v=(a+1,b+1,v)$ to get the orthogonal normal vector to $t$ and $s$.
And use $q\cdot t=0$ and $q\cdot s=0$.
I know this is wrong, and I should be using $N\cdot t=0$ and $N\cdot s=0$, but I don't understand why my logic is wrong. Isn't $q=N-v$ normal to the plane since $v$ is in the plane already?
$q = N - v$ will not be normal to the plane. If N is normal to the plane, and v is a vector in the plane, then q is not normal unless v is zero. Consider a vector in a plane, then also a normal vector to that plane, the difference between the two will be a hypotenuse of a triangle connecting the end of one vector to the end of the normal vector, and that will not be normal since it's not the normal vector. If you want a vector normal to your plane, use the cross product with 2 vectors that are in your plane. Also, q is just a point on your plane, not a vector that coincides with your plane. If you want a vector that is in your plane, subtract one point in your plane from another. e.g. $v - w$.