Understanding Collinear Points and Their Implications on Planes

In geometry, collinear points refer to points that lie on the same straight line. When dealing with the equation of a plane, understanding the collinearity of points is crucial for determining the uniqueness or multiplicity of such planes. Let's explore the scenario where three points are collinear and how it affects the equation of the plane containing them.

Identifying Collinearity

Collinearity can be determined by checking if the vectors formed by any two pairs of the points are scalar multiples of each other. For instance, if you have three points A, B, and C, the vectors (overrightarrow{AB}) and (overrightarrow{AC}) should be proportional for the points to be collinear. This relationship is given by the condition that (overrightarrow{AB} koverrightarrow{AC}) for some scalar (k).

Implications of Collinear Points on Planes

When considering a plane containing three points A, B, and C, the points being collinear means that there is not just one, but infinitely many planes that can contain the line passing through A, B, and C. This is because any plane that contains the line through these points can also contain any point on that line. Let's denote the coordinates of the points A, B, and C as (A(x_1, y_1, z_1)), (B(x_2, y_2, z_2)), and (C(x_3, y_3, z_3)) respectively.

The equation of a plane with general form is (Ax By Cz D 0). However, if points A, B, and C are collinear, then any plane containing the line through these points must satisfy certain conditions. Since the points are on the same line, the normal vector ((A, B, C)) of the plane must be orthogonal to any vector in the plane, including the vector parallel to the line through A, B, and C. This implies that the plane can be any that includes the line, leading to an infinite number of possible planes.

Practical Applications and Examples

Let's consider a specific scenario where we have three points such that the first point A is at (left(frac{1}{21}, frac{30}{21}, frac{3}{21}right)) and the second point B is at ((-2, -3, 5)). We need to check if these points are collinear. First, we convert the fractions to simpler terms: A is at (left(frac{1}{21}, frac{10}{7}, frac{3}{21}right)), which simplifies to approximately A(0.0476, 1.4286, 0.1429). Point B remains (-2, -3, 5).

Calculate the vector (overrightarrow{AB}) which is ((-frac{42}{21}, -frac{70}{21}, frac{105}{21})). Simplifying, this becomes ((-2, -frac{10}{3}, 5)). Now, consider vector (overrightarrow{AC}) where C is at a certain point (let's assume C is at (1, 2, 3) for simplicity in calculation). Then (overrightarrow{AC} (1 - frac{1}{21}, 2 - frac{30}{21}, 3 - frac{3}{21}) (frac{19}{21}, -frac{14}{21}, frac{60}{21})). Simplifying, (overrightarrow{AC} (0.9048, -0.6667, 2.8571)).

Examining these vectors, we see that they do not appear to be scalar multiples of each other, suggesting that points A and B (and C) are not on the same line. Hence, they do not form a collinear set.

However, if you have a collinear set, such as A(0, 0, 0), B(1, 1, 1), and C(2, 2, 2), you can see that the vector (overrightarrow{AB} (1, 1, 1)) and (overrightarrow{AC} (2, 2, 2) 2 overrightarrow{AB}). This confirms collinearity. Since the points are collinear, any plane through this line can be described by a general equation, but the specific equation would depend on the normal vector orthogonal to the line.

Solving for the Plane Equation

Suppose you have a line passing through the points A and B. The direction vector of the line is (overrightarrow{d} (x_2 - x_1, y_2 - y_1, z_2 - z_1)). To find the equation of a plane containing this line and a point P(x?, y?, z?) not on the line, we need a normal vector perpendicular to the direction vector. Let's assume P is (x?, y?, z?).

The normal vector (overrightarrow{n}) can be any vector perpendicular to (overrightarrow{d}) . One way to find such a vector is to choose a vector in the plane that is not parallel to (overrightarrow{d}). For simplicity, let's choose (overrightarrow{n} (0, 0, 1)) if (overrightarrow{d} (1, 0, 0)).

The equation of the plane containing the line and point P would then be in the form:

[0 cdot (x - x_0) 0 cdot (y - y_0) 1 cdot (z - z_0) 0]

Simplifying, we get:

[z - z_0 0]

This plane is defined by the line through the points and a normal vector. If the points are not collinear, the equation is unique and would be derived from a different normal vector that is not parallel to the direction vector of the line.

Conclusion

In summary, if you have collinear points, determining the equation of a plane containing them is not a straightforward process since there are infinitely many such planes. You need to verify the collinearity of the points first and then use a normal vector that is not parallel to the direction vector of the line through the points to find a specific equation for the plane. Understanding these geometric properties is crucial for solving such problems effectively.