Question

Find an equation of the plane. The plane passes through \((2,3,-2),(3,4,2),\) and (1,-1,0) .

Short answer

The equation of the plane is \(8x + 2y + 3z = 16\).

Step-by-step solution

Forming the Vectors

From the given points, form two vectors that lie in the plane. If the points are \(A (2, 3, -2)\), \(B (3, 4, 2)\), and \(C (1, -1, 0)\), we can form the vectors \(AB = B - A = (3-2, 4-3, 2-(-2)) = (1, 1, 4)\) and \(AC = C - A = (1 - 2, -1 - 3, 0 - (-2)) = (-1, -4, 2)\).

Finding the Normal Vector

Next, find the cross product of vectors AB and AC to give a normal vector to the plane. The cross product of vectors AB = (1, 1, 4) and AC = (-1, -4, 2) is (-8, -2, -3). This gives the coefficients of x, y and z in the plane equation.

Forming the Plane Equation

Use the coefficients from the normal vector and one of the points on the plane (for example, A=(2,3,-2)), to find the constant term in the plane equation. So, the plane equation becomes: -8x -2y -3z = d. Plug point A into the equation to solve for d: -8*2 -2*3 -3*(-2). This results in d = -16. Therefore, the equation of the plane is -8x -2y -3z = -16 or expressed in a more standard form: 8x + 2y + 3z = 16.

Key concepts

Vector Cross Product

Understanding the cross product of vectors is pivotal when trying to determine the equation of a plane in three-dimensional space. It starts by taking two vectors that lie on the same plane. In our example, vectors AB = (1, 1, 4) and AC = (-1, -4, 2) are formed from points that the plane passes through. The cross product, symbolized by AB × AC, is a binary operation on two vectors in three-dimensional space and results in a third vector which is perpendicular to the plane containing the first two.(AB) × (AC) = (1, 1, 4) × (-1, -4, 2) = (-8, -2, -3).The resulting vector is the normal to the plane, and its components give us the coefficients for the x, y, and z terms in the plane equation. The cross product is calculated using the determinant of a matrix constructed from the unit vectors i, j, k and the components of the two vectors.

The mathematical representation for the cross product A × B of vectors A = (a1, a2, a3) and B = (b1, b2, b3) is given by the determinant of the following matrix:| i j k || a1 a2 a3 || b1 b2 b3 |. This operation is used not only in finding the plane equation but also in physics for finding torque and in computer graphics for determining the orientation of surfaces.

Normal Vector

A normal vector is fundamental in the geometry of planes. In the context of a plane in three-dimensional space, a normal vector is a vector that is perpendicular to the plane. It essentially defines the orientation of the plane in space. Each plane has an infinite number of normal vectors, but they all are in the same direction or the exact opposite. The cross product of any two non-parallel vectors in the plane will yield a normal vector. In our problem, the vectors AB and AC are used to calculate the cross product, which gives us (-8, -2, -3), a normal vector to the plane.

If a vector AB is represented by two points A and B, its coordinates are simply the differences between the corresponding coordinates of B and A. In vector notation, AB = B - A. The normal vector's relevance is significant since it gives the coefficients directly needed for the plane's equation.

Plane Equation Derivation

Deriving the equation of a plane involves finding a formula that represents all points on that plane. The general form of the plane equation in three dimensions is Ax + By + Cz + D = 0, where A, B, and C are the x, y, and z coefficients derived from a normal vector to the plane, and D is the constant that anchors the plane in space.

To determine D, we use a known point on the plane, which in our case is A (2, 3, -2). Using the normal vector N = (-8, -2, -3) we found earlier, we can plug the x, y, and z values from point A into the equation -8x - 2y - 3z = D to solve for D. After substituting the values, we get D = -16, solidifying the plane equation as 8x + 2y + 3z = 16. The derivation process is vital for translating a set of points and vectors into a comprehensive geometrical representation that describes a unique plane in space.