Plane
In mathematics, a plane is a flat, two-dimensional surface that extends infinitely in all directions
In mathematics, a plane is a flat, two-dimensional surface that extends infinitely in all directions. It is often represented by a sheet of paper or a chalkboard.
A plane is defined by a unique set of three non-collinear points or by an equation. If we have three non-collinear points A, B, and C, we can uniquely determine a plane. This plane is called the plane ABC or plane A.
Alternatively, we can represent a plane using an equation called the general form of the equation of a plane. The general form of a plane equation is ax + by + cz + d = 0, where a, b, and c are constants that represent the direction vector of the plane’s normal, and (x, y, z) are the coordinates of any point on the plane.
To find the equation of a plane given three non-collinear points, we can use the concept of vectors. Let’s say we have the points A(x1, y1, z1), B(x2, y2, z2), and C(x3, y3, z3). We can find two vectors in the plane by subtracting the coordinates of the points: vector AB = B – A and vector AC = C – A.
Next, we find the cross product of the two vectors to get a vector that is orthogonal (perpendicular) to the plane. Let’s call this vector n. The direction vector of the plane’s normal is then the coefficients of n.
Finally, we can substitute one of the points (A, B, or C) and the direction vector of the plane’s normal into the general equation of a plane to find the specific equation of the plane.
For example, let’s find the equation of a plane given the points A(1, 2, 3), B(4, 5, 6), and C(2, 4, 6).
– Find vector AB: AB = B – A = (4 – 1, 5 – 2, 6 – 3) = (3, 3, 3)
– Find vector AC: AC = C – A = (2 – 1, 4 – 2, 6 – 3) = (1, 2, 3)
– Find the cross product of vector AB and AC: n = AB × AC = (3, 3, 3) × (1, 2, 3) = (3, -6, 3)
– The coefficients of n give the direction vector of the plane’s normal: (a, b, c) = (3, -6, 3)
– Choose point A(1, 2, 3) and substitute all values into the general equation of a plane: 3x – 6y + 3z + d = 0
– Solve for d by substituting the coordinates of A: 3(1) – 6(2) + 3(3) + d = 0
– Simplify and solve for d: -12 + d = 0 => d = 12
– The equation of the plane is 3x – 6y + 3z + 12 = 0.
Remember to always check your work and substitute the coordinates of the points into the equation to ensure they satisfy the equation and lie on the plane.
More Answers:
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Determinant Method: Testing Collinearity of Points in Mathematics