Collinearity and Coplanarity of Points

Collinearity of Three Points

Let \( A, B, C \) be three points in space, and let \( O \) be a reference point. The position vectors of \( A, B, C \) with respect to \( O \) are given by:

\[ \overrightarrow{OA}, \quad \overrightarrow{OB}, \quad \overrightarrow{OC}. \]

Then, the points \( A, B, C \) are collinear if and only if there exist scalars \( x, y, z \), not all zero, such that:

\[ x + y + z = 0, \quad x\overrightarrow{OA} + y\overrightarrow{OB} + z\overrightarrow{OC} = 0. \]

Proof

Assume that \( A, B, C \) are collinear. Then \( B \) must divide the segment \( AC \) in some ratio \( m:n \), meaning:

\[ \overrightarrow{OB} = \frac{m\overrightarrow{OC} + n\overrightarrow{OA}}{m+n}. \]

Rearranging,

\[ (m+n)\overrightarrow{OB} = m\overrightarrow{OC} + n\overrightarrow{OA}. \]

Rewriting,

\[ n\overrightarrow{OA} - (m+n)\overrightarrow{OB} + m\overrightarrow{OC} = 0. \]

Setting \( x = n \), \( y = -(m+n) \), and \( z = m \), we obtain:

\[ x + y + z = 0. \]

Thus, the condition holds.

Assume conversely that there exist scalars \( x, y, z \) such that:

\[ x + y + z = 0, \quad x\overrightarrow{OA} + y\overrightarrow{OB} + z\overrightarrow{OC} = 0. \]

Rewriting using \( z = -x - y \),

\[ x\overrightarrow{OA} + y\overrightarrow{OB} - (x + y)\overrightarrow{OC} = 0. \]

Rearranging,

\[ (x + y)\overrightarrow{OC} = x\overrightarrow{OA} + y\overrightarrow{OB}. \]

Dividing by \( x + y \), assuming \( x+y \neq 0 \):

\[ \overrightarrow{OC} = \frac{x\overrightarrow{OA} + y\overrightarrow{OB}}{x+y}. \]

This equation shows that \( C \) divides the segment \( AB \) in the ratio \( y:x \), implying that \( A, B, C \) are indeed collinear.

Thus, the necessary and sufficient condition for three points to be collinear is the existence of scalars \( x, y, z \) such that:

\[ x + y + z = 0, \quad x\overrightarrow{OA} + y\overrightarrow{OB} + z\overrightarrow{OC} = 0. \]

\(\blacksquare\)

Coplanarity of Four Points

Let \( A, B, C, D \) be four points in space, and let \( O \) be a reference point. The position vectors of \( A, B, C, D \) with respect to \( O \) are given by:

\[ \overrightarrow{OA}, \quad \overrightarrow{OB}, \quad \overrightarrow{OC}, \quad \overrightarrow{OD}. \]

Then, the points \( A, B, C, D \) are coplanar if and only if there exist scalars \( \lambda_1, \lambda_2, \lambda_3, \lambda_4 \), not all zero, such that:

\[ \lambda_1 + \lambda_2 + \lambda_3 + \lambda_4 = 0, \]

\[ \lambda_1 \overrightarrow{OA} + \lambda_2 \overrightarrow{OB} + \lambda_3 \overrightarrow{OC} + \lambda_4 \overrightarrow{OD} = 0. \]

Proof

Let \( A, B, C, D \) be coplanar. Suppose they form a planar quadrilateral (if they form a triangle, meaning three points are collinear, or a straight line, then they are already coplanar). Since the four points lie in the same plane, there must be a line joining two of the four points that intersects the line joining the other two points.

Assume that \( AC \) intersects \( BD \) at some point \( P \). Then, \( P \) divides \( AC \) in the ratio \( m:n \) and \( BD \) in the ratio \( p:q \), so its position vector can be written in two ways:

\[ \overrightarrow{OP} = \frac{m \overrightarrow{OC} + n \overrightarrow{OA}}{m+n} \quad \text{or} \quad \overrightarrow{OP} = \frac{p \overrightarrow{OD} + q \overrightarrow{OB}}{p+q}. \]

Equating both expressions,

\[ \frac{m \overrightarrow{OC} + n \overrightarrow{OA}}{m+n} = \frac{p \overrightarrow{OD} + q \overrightarrow{OB}}{p+q}. \]

Cross multiplying,

\[ n(m+n) \overrightarrow{OA} - q(p+q) \overrightarrow{OB} + m(m+n) \overrightarrow{OC} - p(p+q) \overrightarrow{OD} = 0. \]

Setting \( \lambda_1 = \frac{n}{m+n} \), \( \lambda_2 = -\frac{q}{p+q} \), \( \lambda_3 = \frac{m}{m+n} \), \( \lambda_4 = -\frac{p}{p+q} \), we observe that the sum of all coefficients is:

\[ \lambda_1 + \lambda_2 + \lambda_3 + \lambda_4 = 0. \]

Thus, the required condition holds.

Conversely, suppose there exist scalars \( \lambda_1, \lambda_2, \lambda_3, \lambda_4 \) such that:

\[ \lambda_1 + \lambda_2 + \lambda_3 + \lambda_4 = 0, \quad \lambda_1 \overrightarrow{OA} + \lambda_2 \overrightarrow{OB} + \lambda_3 \overrightarrow{OC} + \lambda_4 \overrightarrow{OD} = 0. \]

Rearranging,

\[ \lambda_1 + \lambda_2 = -(\lambda_3 + \lambda_4), \quad \lambda_1 \overrightarrow{OA} + \lambda_2 \overrightarrow{OB} = - (\lambda_3 \overrightarrow{OC} + \lambda_4 \overrightarrow{OD}). \]

Dividing both sides by \( \lambda_1 + \lambda_2 \),

\[ \frac{\lambda_1 \overrightarrow{OA} + \lambda_2 \overrightarrow{OB}}{\lambda_1 + \lambda_2} = - \frac{\lambda_3 \overrightarrow{OC} + \lambda_4 \overrightarrow{OD}}{\lambda_3 + \lambda_4}. \]

Since the left-hand side represents a point dividing \( AB \) in the ratio \( \lambda_2 : \lambda_1 \) and the right-hand side represents a point dividing \( CD \) in the ratio \( \lambda_4 : \lambda_3 \), this implies that \( AB \) and \( CD \) intersect, proving that \( A, B, C, D \) are coplanar.

At first, defining collinearity and coplanarity using vectors may seem unusual. We are used to thinking of points being on the same line or plane by looking at diagrams. However, expressing these conditions with position vectors helps us describe them algebraically in a way that does not depend on specific coordinates or visual intuition.

This approach forms the foundation for many important concepts. While this may initially seem abstract, it is a powerful way of thinking that will unify many key ideas in geometry and vectors.