Question

In Exercises \(41-52,\) find an equation of the plane. The plane passes through \((0,0,0),(1,2,3),\) and (-2,3,3) .

Short answer

The equation of the plane that passes through the points (0,0,0), (1,2,3), and (-2,3,3) is -3x -y + 5z = 0.

Step-by-step solution

Find Vectors

Create two vectors from the three given points. Let vector AB = B - A and vector AC = C - A, where A = (0,0,0), B = (1,2,3) and C = (-2,3,3). This gives us AB = (1-0, 2-0, 3-0) = (1,2,3) and AC = (-2-0, 3-0, 3-0) = (-2,3,3).

Calculate Cross Product

The cross product of AB and AC gives a vector that is perpendicular (or normal) to the plane. Use the formula for cross product between two 3D vectors to find it. AB x AC = \(i( 2 * 3 - 3 * 3) - j( 1 * 3 - 3 * -2) + k(-2 - 3 * 1)\). Simplify to get AB x AC = \(- 3i - 1j + 5k\).

Form the equation of the plane

The equation of a plane can be written as Ax+By+Cz=D, where A, B and C are the i, j, k components of a normal vector to the plane, (x, y, z) are coordinates of any point on the plane, and D is a constant. From the normal vector \(-3i - 1j + 5k\), A = -3, B = -1, C = 5. Plug in A, B, C, and the coordinates of any given point (for instance, the origin) into the equation to get D. Then form the equation of the plane with those values.