Given the following plane equation. x + y + z = 1 and three points A(1, 0, 0), B

Question

Given the following plane equation. x + y + z = 1 and three points A(1, 0, 0), B(0.5, 0.3, 0.1), C(0.2, 0.5, 0.8), decide their locations relative to the given plane, i.e., in front of the plane? on the plane? or behind the plane? One point was originally located at (100, 100). Suppose we move it for 20 units along positive x-axis and then 30 units along positive y-axis. Then rotate the point counter- clockwise 30 degrees about the coordinate origin. Use the homogeneous coordinates to compute the new coordinates for this point after translation and rotation.

Explanation / Answer

Method: Substitute the points in the LHS of equation of the plane (x + y + z). If the value is equal to RHS, ie 1, then the point is on the plane, if value is less than RHS, then point is below the plane, and if value is greater than RHS, then point is above the plane.

A(1,0,0). x = 1, y = 0, z = 0. LHS is x+y+z = 1+0+0 = 1. Which is equal to RHS. Hence point A is on the plane.

B(0.5,0.3,0.1). x= 0.5, y = 0.3, z = 0.1. LHS is x+y+z = 0.5 + 0.3 + 0.1 = 0.9 < 1. Hence point B is below the plane.

C(0.2,0.5,0.8). x = 0.2, y = 0.5, z = 0.8. LHS is x+y+z = 0.5+0.2+0.8 = 1.5 > 1. Hence point C is above the plane.

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