Coplanar points

If there is a plane that contains every point of a given set, those points are called coplanar.

We'll be using vectors and specifically the cross product and dot product.

We want to check if the points \(A,\) \(B,\) \(C\) and \(D\) are coplanar.

First consider the plane \(ABC.\) The cross product of the vectors \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\) is normal to the plane they are on.

Therefore, if the vector \(\overrightarrow{AD}\) is normal to \(\overrightarrow{AB} \times \overrightarrow{AC},\) all four points are on the same plane. That is, we want to prove:

\[ \left(\overrightarrow{AB} \times \overrightarrow{AC}\right) \cdot \overrightarrow{AD} = 0 .\]

Are \( A =(1, 2, 2),\) \(B=(3, -2, 4),\) \( C= (5, -1, 5),\) and \(D = (3, 1, 3)\) on the same plane?

First we want to find the cross product \( \overrightarrow{AB} \times \overrightarrow{AC} .\) The vectors are \( \overrightarrow{AB} = (3-1)i + (-2-2)j + (4-2)k = 2i -4j + 2k \) and \( \overrightarrow{AC} = (5-1)i + (-1-2)j + (5-2)k = 4i - 3j + 3k .\)

The cross product is then \( \left((-4)(3)-(2)(-3)\right)i+\left((2)(4)-(2)(3)\right)j +\left((2)(-3)-(-4)(4)\right)k = -6i + 2j + 10k .\)

The vector \( \overrightarrow{AD} \) is \( (3-1)i + (1-2)j + (3-2)k = 2i -1j +1k .\) We then just need the dot product of this with \(-6i + 2j + 10k.\)

\[ (2)(-6) + (-1)(2) + (1)(10) = -12 - 2 + 10 = -4\]

Since \(-4 \neq 0 ,\) the four points are not coplanar.

The plane containing \( A, B,\) and \(C,\) with \(D\) marked in blue.