Example on Section 12.5: Planes in Space

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Example on Section 12.5: Planes in Space

In general, a plane in space is determined by specifying a vector that is normal to the plane (i.e., orthogonal to every vector lying in the plane) and a point P1(x1, y1, z1) lying in the plane (see Figure above). In order to find an equation of the plane, let P(x, y, z) represent any point in the plane. Then, since P and P1 are both points in the plane, the vector lies in the plane and so, must be orthogonal to a. Their dot product is then zero. So we have the vector equation of a plane:
Which means;
a1(x-x1) +a2(y-y1) +a3(z-z1)=0, or a1x+a2y +a3z=d with d= a1x1+ a2y1+ a3z1. Scalar equations
These equations are for the plane passing through the point P1(x1, y1, z1) with normal vector As you can imagine, it's a simple matter to use this to find the equation of any particular plane. We illustrate this in the following example.
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/ The Equation of a Plane Given a Point and a Normal Vector
Find an equation of the plane containing the point (1, 2, 3) with normal vector

The scalar equation is 4(x-1)+5(y-2)+6(z-3)=0. (E), or by simplification: 4x+5y+6z=32
To draw the plane, we locate three points lying in the plane. In this case, the simplest way to do this is to look at the intersections of the plane with each of the coordinate axes.
When y = z = 0, we get from (E) that 4(x-1)+5(0-2)+6(0-3) = 4x-4-10-18=0so that 4x = 32 or x = 8.
The intersection of the plane with the x - axis is then the point (8, 0, 0).
Similarly, you can find the intersections of the plane with the y- and z - axes: and respectively. Using these three points, we can draw the plane seen above. While we have only drawn a triangle, you can easily visualize the plane we are interested in by extending the sides of the triangle. Notice that since the plane intersects all three of the coordinate axes, all that we see in the first octant is a triangle.
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/ Finding the Equation of a Plane Given Three Points
Find the plane containing the three points P(1, 2, 2), Q(2, -1, 4) and R(3, 5, -2).

First, we'll need to find a vector normal to the plane (any one). Notice that two vectors lying in the plane are A vector orthogonal to both is the cross product

must be orthogonal to the plane: it is the normal vector a (or n) we are looking for. The equation for the plane with this normal vector and containing P(1,2,2) is then
0 = 6(x-1)+8(y-2)+9(z-2) or 6x+8y+9z=40
Angle between two planes: In three dimensions, two planes are either parallel or they intersect in a straight line. (Think about this some.) Suppose that two planes having normal vectors a and b, respectively, intersect. Then the angle between the planes is the same as the angle between a and b
Angle between planes.
With this in mind, we say that the two planes are parallel whenever their normal vectors are parallel and the planes are orthogonal whenever their normal vectors are orthogonal.
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/ The Equation of a Plane Given a Point and a Parallel Plane
Find an equation for the plane through the point (1, 4, -5) and parallel to the plane 2x-5y+7z = 12.

First, notice that a normal vector to the given plane is Since the two planes are to be parallel, this vector is also normal to the new plane. So we can write down the equation of the plane:
2x-5y+7z= d where d= 2(1)-5(4)+7(-5)= 53
/ It's particularly easy to see that some planes are parallel to the coordinate planes.
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/ Drawing Some Simple Planes
Draw the plane y = 3; draw the plane y = 8.

First, notice that both equations represent planes with the same normal vector, This says that the planes are both parallel to the xz - plane, the first one passing through the point (0, 3, 0) and the second one passing through (0, 8, 0).
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/ Finding the Intersection of Two Planes
Find the line of intersection of the planes: x+2y+z = 3 and x-4y+3z = 5.
You should recognize that the intersection of two nonparallel planes will be a line. (Think about this some!) In this example, we see how to find this line.

Solving both equations for x, we get x=3-2y-z and x=5+4y-3z (E1)
Setting these expressions for x equal gives us 3-2y-z =5+4y-3z
Solving this for z gives us 2z=6y+2 or z=3y+1
Returning to either equation in (E1), we can solve for x (also in terms of y ). We have
x = 3-2y-z = 3-2y-(3y+1) = -5y+2. So x=-5y+2
If we take y as our parameter (if you wish, let y = t), we have parametric equations for the line of intersection:

You can see the line of intersection in the computer-generated graph of the two planes
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Suppose that we wanted to find the distance from the planeax+by+cz+d=0 to a point P0(x0, y0, z0) not on the plane. Notice that the distance is measured along a line segment connecting the point to the plane that is orthogonal to the plane (see Figure). To compute this distance, pick any point P1(x1, y1, z1) lying in the plane and let a = denote a vector normal to the plane. Next, notice from the Figure that the distance from P0 to the plane is simply . Also notice that

From section 12.3 (see section12.5 example8), we can now write the distance as

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/ Finding the Distance between Parallel Planes
Find the distance between the parallel planes: P1: 2x-3y+z=6 and P1: 4x-6y+2z=8

First, notice that the planes are parallel, since their normal vectors and are parallel (one is twice the other or check that there cross product is zero). Next, notice that since the planes are parallel, the distance from the plane P1 to every point in the plane P2 is the same. So, pick any point in P2, say (0, 0, 4); (This is certainly convenient.) The distance d from the point (0, 0, 4) to the plane P1 is then given by