3D Vectors Math Problem o.o?
How do you find the point where the line joining (2,3,1) to (4,-2,5) cuts the plane of 2x+y-z=3?
I'm assuming you got to make the two points into a vector, which would be (2,5,6) but then what do you do?
Thank in advance homies 
Comments
You need to write the equation of the line going through your two points, and then find the point that is both on that line and on the plane, i.e. solution of the system of the two equations.
You can write a line in two different ways. The Catersian and parametric ways. When you have two points (like you have) or one point and a vector (like you almost have) the easiest way is the parametric line. Point P0 (2,3,1) and V0 (4-2,-2-3,5-1) = (2,-5,4)
X = V0x *t + P0x = 2t+2
Y = V0y *t + P0y = -5t+3
Z = V0z *t + P0z = t+1
The equation of your plane is 2x+y-z=3.
So you find the parameter t such that the point of the line (X,Y,Z) is also on the plane by plugging X,Y,Z into your plane equation and solving for t. You get:
2( 2t+2 )+( -5t+3 )-( t+1 ) = 3
4t+4 -5t+3 - t-1 ) = 3
-2t +6 = 3
t = 3/2
So the solution is the point ( 2(3/2)+2, -5(3/2)+3, (3/2)+1 ) = ( 5, -9/2, 5/2 ).
Write the equations of the line thr these points. They are
(x-2)/2 + (y-3)/-5 = (z-1)/4 = t say so x= (2t+2) y = (-5t+3) and z = (4t + 1)
for any point on the line. Now find t by substituting thesex, y and z in the equation of the plane, find t and then x, y, z.
What vector you have typed is incorrect.
Why did you do the bypass product? mindset of A is arctan(a million) = 45º. it should be in Q2, so the perspective is 100 80–40 5 = 100 thirty fiveº perspective of B is arctan(2/5) = 21.8º mindset between both is 100 thirty 5–21.8= 113.2º angles measured CCW from +x axis