98 CHAPTER 6. VECTOR PRODUCTS

The two normal vectors are (1,2,−1) and (3,2,−1) . Therefore, the cosine of the angledesired is

cosθ =(1,2,−1) · (3,2,−1)√

12 +22 +(−1)2√

32 +22 +(−1)2= .87287

Now use a calculator or table to find what the angle is. cosθ = .87287, Solution is :{θ = .50974} . This value is in radians.

Sometimes you need to find the equation of a plane which contains three points. Con-sider the following picture.

(a0,b0,c0)

(a1,b1,c1)

(a2,b2,c2)

a

b

You have plenty of points but you need a normal. This can be obtained by taking a×bwhere a= (a1−a0,b1−b0,c1− c0) and b= (a2−a0,b2−b0,c2− c0) .

Example 6.7.5 Find the equation of the plane which contains the three points

(1,2,1) ,(3,−1,2) ,and (4,2,1) .

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

(3,−1,2)− (1,2,1) and (4,2,1)− (1,2,1)

Thus a normal vector is (2,−3,1)×(3,0,0)= (0,3,9) . Therefore, the equation of the planeis

0(x−1)+3(y−2)+9(z−1) = 0

or 3y+ 9z = 15 which is the same as y+ 3z = 5. When you have what you think is theplane containing the three points, you ought to check it by seeing if it really does containthe three points.

Example 6.7.6 Find the equation of the plane which contains the three points

(1,2,1) ,(3,−1,2) ,and (4,2,1) .

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

(3,−1,2)− (1,2,1) and (4,2,1)− (1,2,1)

Thus a normal vector is (2,−3,1)×(3,0,0)= (0,3,9) . Therefore, the equation of the planeis

0(x−1)+3(y−2)+9(z−1) = 0

or 3y+9z = 15 which is the same as y+3z = 5.