I am trying to calculate the normal vector of a given plane with dot products on the one hand(Attempt 1) and with vector cross products on the other (Attempt 2). The results aren't making any sense to me: shouldn't both attempts which result in orthogonality to vectors in the same plane be similar or related? By both having orthogonal vectors I don't mean orthogonal results. The dot product has a scalar result and the cross product a vector. But I solve for the vector that produces the scalar result in Attempt 1.
Points/Vectors Setup:
Same plane, points normalized:
pt1 (-2.562, 571.286, 1)
pt2 (1014.394, 577.530, 1)
pt3 (-4.972, -3.807, 1)
Let $\vec u =~$pt2-pt1 $= ~(1016.956,6.244, 0)$.
Let $\vec v = $unknown orthogonal vector for Method 1.
Let $\vec r = $pt3-pt1 $(-2.41,-575.093, 0)$
Let $\vec n =~$unknown orthogonal vector to $\vec u ,\vec v$.
Attempt 1: With Dot Product Find $~\vec v~$ Orthogonal to Plane Vector $\vec u$:
So,
$\vec u\cdot \vec v = 0~$ if $~ \vec u ⟂\vec v $.
Solve for $\vec v~ $unknown.
$\vec u\cdot \vec v = (u_x,u_y,u_z)\cdot(v_x, v_y, v_z)~$
$~~~~~~~=(u_xv_x+u_yv_y+u_zv_z)$
$\vec u = (u_x, u_y, u_z)$
$u_x = 1016.956, u_y =6.244 , u_z = 0$ from points/vectors above
$\vec u\cdot \vec v =(1016.956v_x+6.244v_y+0v_z)$
Simplify equation, $v_z$ component = 0:
$\vec u\cdot \vec v =(1016.956v_x+6.244v_y)$
Set $\vec u\cdot \vec v = 0$, our unknown $(\vec v )$will be perpendicular to $\vec u$
Choose values that complete the equation:
Select y = 10, while z = 0 already, leaving $v_x= -6.244(10)/1016.956 =-0$.0614
$\therefore$ unknown$~\vec v = (-.0614, 10, 0$).
Attempt 1: Test the Answer:
Is $\vec v ⟂ \vec u?$
if true $\vec v \cdot \vec u = 0$
ok, check:
$\vec v \cdot \vec u = (1016.956, 6.244, 0)\cdot(-.0614, 10, 0)$
$~~~~~~~ =-1.108$
$~~~~~~\therefore ~\vec v ⟂ \vec u.$
Attempt 2: Find $\vec n$ from cross product
Using two vectors $\vec r, \vec u~$ from the same plane:see(from Points/Vectors)
$\vec u × \vec r = \vec n$
$\vec u × \vec r = (1016.956, 6.244, 0)×(-2.41,-575.093, 0)$
$\vec n = (0,0, -584829.229)$
Are the 2 vectors orthogonal?:
$\vec n\cdot\vec u = 0$?
$(0,0,-584829.229)\cdot (1016.956, 6.24, 0) = 0,~\therefore ~$yes.
check if $\vec n \cdot \vec v = 0$?
$(0,0,-584829.229)\cdot (-2.41,-575.093, 0) = 0,~\therefore ~$yes.
Best Answer
Your first solution is incomplete. You’ve found one vector that’s perpendicular to $\vec u$, but it also needs to be perpendicular to $\vec r$. Even with the correct value for $\vec u$, choosing $\vec v$ in the way you did doesn’t produce a vector that’s perpendicular to $\vec r$. You need to solve the system of equations $\vec u\cdot\vec v=0$, $\vec r\cdot\vec v=0$.
At the risk of confusing things, there’s another rather direct method available that you didn’t try. The equation of the plane is going to have the form $ax+by+cz+d=0$. If you plug in the coordinates of the three points, you’ll get a system of three linear equations in the unknown coefficients $a$, $b$, $c$ and $d$. Solve that system.