Generalize and show that there is a plane ax + by + cz + d = 0 through any three points in space
I get that we have two equations ax+by+cz+d=0 and ax1+by1+cz1+d=0 but from then on I have no idea what to do thanks
Generalize and show that there is a plane ax + by + cz + d = 0 through any three points in space
I get that we have two equations ax+by+cz+d=0 and ax1+by1+cz1+d=0 but from then on I have no idea what to do thanks
Consider 3 points P1,P2,P3 with coordinates x1,2,3 , y1,2,3 and z1,2,3 .............
consider the vectors P1P2 and P1P3 . then form the normal vector n = (P1P2 X P1P3)
to define the unique plane that passes through P1,P2,P3..... just get the scalar product between the normal vector and any vector P1P P(x,y,z) of the pane. You will obtain an equation ax+by+cz =d which is the equation of the plane you are looking for....
You are having three points that will give you three equations
Equation of a plane through a point say ( x1, y1, z1) is given by
a(x-x1)+b(y-y1)+ c(z-z1) = 0
Now it passes through two other points ( x2, y2, z2 ) and ( x3, y3, z3) Thus we have
a(x2-x1)+b(y2-y1)+ c(z2-z1) = 0 and
a(x3-x1)+b(y3-y1)+ c(z3-z1) = 0
solve these equations for a,b and c we get
a/---- = b/ ----- = c/ ----- etc. put it equal to say P
Now you have a,b and c in terms of P
Plug theses values in any one of the two equations to get P
and you have the required equation of the plane.