Normals and Planes

Let's suppose that n = < a,b,c > is a nonzero vector and that P[1] ( x[1], y[1], z[1] ) is a point in space. We then define the plane with normal n through the point P[1] to be the set of points P ( x,y,z ) for which the vectors

v = conjugate(P[1]*P) = < x-x[1] , y-y[1] , z-z[1] >

are perpendicular to n .

(commands to generate the figure below)

[Maple Plot]

The equation of the plane follows from the fact that n perpendicular to v implies that their dot product is zero. ( A period is used to represent a dot product )

n . v = 0 ==> < a,b,c > . < x-x[1] , y-y[1] , z-z[1] > = 0

Thus, the equation of the plane with normal n through the point ( x[1] , y[1] , z[1] ) is given by

a ( x-x[1] ) + b ( y-y[1] ) + c ( z-z[1] ) = 0

Example: Find and plot the equation of the plane through P[1] (3,2,1) with normal n = <1,-1,2>.

Solution: The equation of the plane is

1( x -3) - 1( y -2) + 2( z -1) = 0

x- 3- y+ 2+2 z -2 = 0

x - y + 2 z = 3

In order to plot this equation, we must solve for z :

z = - x + y + 3

We then use "plot3d" to plot the expression on the right side of the equation:

> plot3d(-x+y+3,x=0..4,y=0..4,grid=[5,5],color=red,axes=normal);

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