6.7. PLANES 99

Proposition 6.7.7 If (a,b,c) ̸= (0,0,0) , then ax+ by+ cz = d is the equation of a planewith normal vector ai+bj+ ck. Conversely, any plane can be written in this form.

Proof: One of a,b,c is nonzero. Suppose for example that c ̸= 0. Then the equationcan be written as

a(x−0)+b(y−0)+ c(

z− dc

)= 0

Therefore,(0,0, d

c

)is a point on the plane and a normal vector is ai+bj+ck. The converse

follows from the above discussion involving the point and a normal vector. ■

Example 6.7.8 Find the equation of the plane containing the points (1,2,3) and the line(0,1,1)+ t (2,1,2) = (x,y,z).

There are several ways to do this. One is to find three points and use the above pro-cedures. Let t = 0 and then let t = 1 to get two points on the line. This yields the threepoints (1,2,3) ,(0,1,1) , and (2,2,3) . Then a normal vector is obtained by fixing a pointand taking the cross product of the differences of the other two points with that one. Thusin this case, fixing (0,1,1) , a normal vector is

(1,1,2)× (2,1,2) = (0,2,−1)

Therefore, an equation for the plane is

0(x−0)+2(y−1)+(−1)(x−3) = 0

Simplifying this yields2y+1− x = 0

Example 6.7.9 Find the equation of the plane which contains the two lines, given by thefollowing parametric expressions in which t ∈ R.

(2t,1+ t,1+2t) = (x,y,z) , (2t +2,1,3+2t) = (x,y,z)

Note first that you don’t know there even is such a plane. However, if there is, you couldfind it by obtaining three points, two on one line and one on another and then using any ofthe above procedures for finding the plane. From the first line, two points are (0,1,1) and(2,2,3) while a third point can be obtained from second line, (2,1,3) . You need a normalvector and then use any of these points. To get a normal vector, form (2,0,2)× (2,1,2) =(−2,0,2) . Therefore, the plane is−2x+0(y−1)+2(z−1) = 0. This reduces to z−x = 1.If there is a plane, this is it. Now you can simply verify that both of the lines are really inthis plane. From the first, (1+2t)−2t = 1 and the second, (3+2t)− (2t +2) = 1 so bothlines lie in the plane.

One way to understand how a plane looks is to connect the points where it interceptsthe x,y, and z axes. This allows you to visualize the plane somewhat and is a good way tosketch the plane. Not surprisingly these points are called intercepts.

Example 6.7.10 Sketch the plane which has intercepts (2,0,0) ,(0,3,0) , and (0,0,4) .