![]() The basic idea is that the vector \mathbf v to be reflected is decomposed as \mathbf v\parallel \mathbf v\perp, its respective components parallel to and perpendicular to the reflector. ![]() Thus there are two possible orientations of the sides of the top face of the cube.įinally, the normal to the top cube face is along the cross product of $Q_2$ and $Q_4$, and will thus have two possible normals. Instead of relying on found or memorized formulas, try to work this out for yourself from the fundamentals. The positive solution corresponds to two possible values of $z_2$ (the second being the negative of the first), and from the second original equation we obtain a single unique value of $z_4$ for each of the solutions of $z_2$. Which is a simple quadratic equation in $z_2^2 $ If $Q_1, Q_2, Q_3, Q_4$ are the actual points in $3$D space, then So in this example, AoI(6,(10:12)) 1 I consider a plane in row 6, column 10-12. This plane is a plane considered where the targets are. Now attach a 3D coordinate frame having its $x$ and $y$ axes aligned with the $x$ and $y$ axes of the image, while the $z$ axis is pointing out of the image plane, and its origin at $P_1$. In order to calculate the incident angle between transmitter-target, I need to define a normal vector to this target, a normal vector to this plane. So n dot this vector is going to be equal to 0. So I want to figure out at any given point a vector that's popping straight out in that direction. You take their dot product- it's going to be equal to zero. Obviously you can figure out a normal vector you can just divide it by its magnitude and you will get the unit normal vector. ![]() And this is, this right, the normal vector is normal to the plane. #Normal vector 2d geometry plus#Suppose you have the four image points $P_1, P_2, P_3, P_4$ in the $xy$ plane (the plane of the image) arranged in counterclockwise direction (so that $P_1 P_3$ and $P_2 P_4$ are the two diagonals. It's going to be x minus xpi plus y minus ypj plus z minus zpk. ![]()
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