If The Rocket And The Spy Plane Become Enmeshed In A Tangled Mess Where Relative To The Point Of

For a Rocket to reach an orbit around the Earth, it should have a velocity of ~7,8 km/s i.e., Circular Orbital Velocity.The rocket has to start from rest position and reach this velocity.But you have to notice that the whole time, the Earth is rotating.This rotation is from West to East.At the equator, the velocity is rotation of the Earth is ~ 450 m/s. This gives us an inertia. Thus, to make use of this, most rockets fly East.Example: Soyuz trajectory to ISSThanks - Russian Space WebIf we launch towards West, we have to overcome this 450 m/s. That means more rocket mass and therefore more cost.That said, there are some rockets which fly West. Most of them are launched by Israel because any Israeli rocket flying East would fly over hostile territory. This would be perceived as aggression.

A plane in 3D is defined by 3 parameters (a, b, c) as the set of points which satisfies ax+by+cz=1 (take notice that (a, b, c) represents the normal vector to the plane, it's very important, and that its length determines the offset of the plane from the origin.). Basically your problem can be described as trying to find the best parameters (a, b, c), given that you know 4 points. This is equivalent to trying to solve the system of homogeneous equations[math] ax_1 + by_1 + cz_1 = 1 [/math][math] ax_2 + by_2 + cz_2 = 1 [/math][math] ax_3 + by_3 + cz_3 = 1 [/math][math] ax_4 + by_4 + cz_4 = 1 [/math]which can also be written using matrix notation as [math] PA = 1 [/math], where 1 is a vector of ones of the appropriate dimension.The problem then becomes an optimization problem where you try to minimize [math] \min \| PA - 1 \|^2 [/math]. This has a very simple closed form solution: [math] A = (P^TP)^{-1}P^T1[/math].You can now start to use this method for a generic set of 4 points. In the specific case which you presented in the question, i.e. where P1 = (0, 0, 0), P2 = (10, 0, 0), P3 = (0, 10, 0), P4 = (10 ,10 ,0.1), you will have that the best parameters are (a, b, c) = (.1, .1, -10). Thus the optimal plane is .1x + .1y -10z = 1.If you want to also "fix" your points such that they lie in this plane, then you can do this simply by taking each of them individually, projecting them on the normal of the plane (.1, .1, -10), and then subtracting this projection from the original point. The projection of a point on a vector is [math] Proj(p, n) = \frac{nn^T}{n^Tn}p [/math], and if we subtract this from p we get the result which is [math] p - Proj(p, n) [/math] [math] = (I-\frac{nn^T}{n^Tn}) p [/math]. This result is a point which at the same time lies on the plane and has minimal distance to the original point. Thus you just "fix" your previous point by substituting with this one.Let's do this for point P2 = (10, 0, 0). The normal is n = (a, b, c) = (.1, .1, -10), thus we get that [math] (I-\frac{nn^T}{n^Tn}) p = (9.999, -.0009998 , .09998) [/math] [math] \approx (10, -.001, .1) [/math].

Plane has equation: Ax + By + Cz + D = 0 where vector N=(A,B,C) is the normal (perpendicular) to the plane To find vector N, find 2 points P,Q on perpendicular line: t=0, P=(1,0,2) t=1, Q=(2,2,-1) N = (2-1, 2-0, -1-2) = (1,2,-3) Plane: 1x + 2y - 3z + D = 0 Find D by substituting in values of point (-2,7,10) 1(-2) + 2(7) - 3(10) + D = 0 -2 + 14 - 30 + D = 0 -18 + D = 0 D = 18 Plane: x + 2y - 3z + 18 = 0Understand more about Equation of a Plane