If we take A(3,4,-2) as a reference vector then we can work out the vector difference of B, C, D from that vector: B-A=(5,1, 2); C-A=(-2,6,-4); D-A=(6,-2,4).
The principle behind the solution is that the vector product of two vectors produces a normal vector. If we find the dot product of this result with a third vector and the result is zero, the third vector must be perpendicular to the normal, implying it is coplanar with the other two vectors. Three points are always coplanar because they form a triangle which is a plane object. We calculate vectors from the points by subtraction.
We need to work out some vector products: first the vector product of (5,1,2) and (-2,6,-4). To do this we need to remember:
| |
i |
j |
k |
| i |
0 |
k |
-j |
| j |
-k |
0 |
i |
| k |
j |
-i |
0 |
where i, j and k are unit vectors for x,y and z.
We get two products contributing to values in the unit vectors: i: jk-kj; j: ki-ik; k: ij-ji.
Applying these to the coefficients we get (-4-12,-4+20,30+2)=(-16,16,32).
Now we work out the dot product with (6,-2,4): -96-32+128=0. Zero implies that all the points are coplanar.
The determinant below is another way of expressing the vector (cross) product:
| i j k |
| 5 1 2 | = (-4-12)i-(-20+4)j+(30+2)k=(-16,16,32)
| -2 6 -4 |