You have an idea of what a plane is already. It is the span of some vectors. However, it can also be considered geometrically in terms of a dot product. To find the equation of a plane, you need two things, a point contained in the plane and a vector normal to the plane. Let p_{0} =

whenever p is the position vector of a point in the plane. The following picture illustrates the geometry of this idea.
Expressed equivalently, the plane is just the set of all points p such that the vector p − p_{0} is perpendicular to the given normal vector n.
Example 3.7.1 Find the equation of the plane with normal vector n =
From the above, the equation of this plane is just

Example 3.7.2 2x + 4y − 5z = 11 is the equation of a plane. Find the normal vector and a point on this plane.
You can write this in the form 2
Definition 3.7.3 Suppose two planes intersect. The angle between the planes is defined to be the angle which is ≤ π∕2 than between normal vectors to the respective planes.
Example 3.7.4 Find the angle between the two planes x + 2y − z = 6 and 3x + 2y − z = 7.
The two normal vectors are

Now use a calculator or table to find what the angle is. cosθ = .87287, Solution is :
Sometimes you need to find the equation of a plane which contains three points. Consider the following picture.
You have plenty of points but you need a normal. This can be obtained by taking a × b where a =
Example 3.7.5 Find the equation of the plane which contains the three points

You just need to get a normal vector to this plane. This can be done by taking the cross products of the two vectors

Thus a normal vector is

or 3y + 9z = 15 which is the same as y + 3z = 5. When you have what you think is the plane containing the three points, you ought to check it by seeing if it really does contain the three points.
Example 3.7.6 Find the equation of the plane which contains the three points

You just need to get a normal vector to this plane. This can be done by taking the cross products of the two vectors

Thus a normal vector is

or 3y + 9z = 15 which is the same as y + 3z = 5.
Proposition 3.7.7 If
Proof: One of a,b,c is nonzero. Suppose for example that c≠0. Then the equation can be written as

Therefore,
Example 3.7.8 Find the equation of the plane containing the points
There are several ways to do this. One is to find three points and use the above procedures. Let t = 0 and then let t = 1 to get two points on the line. This yields the three points

Therefore, an equation for the plane is

Simplifying this yields

Example 3.7.9 Find the equation of the plane which contains the two lines, given by the following parametric expressions in which t ∈ ℝ.

Note first that you don’t know there even is such a plane. However, if there is, you could find it by obtaining three points, two on one line and one on another and then using any of the above procedures for finding the plane. From the first line, two points are
One way to understand how a plane looks is to connect the points where it intercepts the x,y, and z axes. This allows you to visualize the plane somewhat and is a good way to sketch the plane. Not surprisingly these points are called intercepts.
Example 3.7.10 Sketch the plane which has intercepts
You see how connecting the intercepts gives a fairly good geometric description of the plane. These lines which connect the intercepts are also called the traces of the plane. Thus the line which joins
Example 3.7.11 Identify the intercepts of the plane 3x − 4y + 5z = 11.
The easy way to do this is to divide both sides by 11. Thus