The Cartesian equation of a plane #P# is #ax + by + cz + d = 0#, where #a, b, c# are the coordinates of the normal vector #vec n = ( (a), (b), (c) ) #
Let #A, B and C# be three noncolinear points, #A, B, C in P#
Note that #A, B and C# define two vectors #vec (AB)# and #vec (AC)# contained in the plane #P#. We know that the cross product of two vectors contained in a plane defines the normal vector of the plane.
Now, we can use an example to illustrate the solution. Assume that the coordinates of the three points are the following:
#A(1,2,3), B(-2,1,0)# and #C(0,3,2)#
From the coordinates of the points #A, B# and #C# we can find the vectors #vec (AB)# and #vec (AC)#:
#vec (AB) = (x_b - x_a)*hat i + (y_b - y_a)*hat j + (z_b - z_a)*hat k#
#vec (AC) = (x_c - x_a)*hat i + (y_c - y_a)*hat j + (z_c - z_a)*hat k#
where #hat i, hat j# and #hat k# are the unit vectors on the cartesian axes of coordinates #Ox, Oy# and #Oz#.
After plugging in the values of the coordinates, we have:
#vec (AB) = (-2 - 1)*hat i + (1-2)*hat j + (0-3)*hat k#
So, #vec (AB) = -3 hat i - hat j -3 hat k#
#vec (AC) = (0-1)*hat i + (3-2)* hat j + (2-3)*hat k#.
So, #vec (AC) = -hat i + hat j - hat k#
Next, we'll find the normal vector from the cross product of #vec (AB)# and #vec (AC)#
#vec n = vec (AB)# x #vec (AC) = | (hat i,hat j,hat k),(-3,-1,-3),( -1,1,-1)|# #=# # 4 hat i - 4 hat k#
Therefore, the equation of the plane is #4x-4z+d=0#. In order to find #d#, we can plug in the coordinates of point #A#:
#d= 4z - 4x => d = 4*3 - 4*1 = 8#
So, the cartesian equation of the plane is #4x-4z+8=0#.
