MCV4U1-UNIT EIGHT-LESSON FIVE
Lesson Five: The Cartesian (or Scalar) Equation of a Plane
We can find the scalar equations of a plane using dot product, just as we did for a line:
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n   A, B, C 
Px, y, z 
P1x1, y1, z1 
Plane
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Let n   A, B, C  be normal to the plane.
Let Px, y, z  be any point on the plane
Let P1x1, y1, z1  be a specific point on the plane.
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n  P1P  0
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n   A, B, C    x  x1 , y  y1 , z  z1 
Ax  Ax1  By  By1  Cz  Cz 1  0
Let D   Ax1  By1  Cz1
Then the scalar equation of a plane is .....
Ax  By  Cz  D  0
Example 1: Find the scalar equation of the plane through the point (1,-1,0) and having normal (2,-1,5).
Solution:
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n  P1P  0
2,1,5  x  1, y  1, z   0
2 x  2  y  1  5z  0
2 x  y  5z  3  0
Example 2: Find the scalar equation of the plane through the point (3,-1,2) and with direction vectors (2,1,0) and (3,5,2).
Solution:
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n  2,1,0 3,5,2  2,4,13
2,4,13  x  3, y  1, z  2  0
2 x  6  4 y  4  13z  26  0
 2 x  4 y  13z  16  0