In the past, when we where studying lines in the plane, we described how a line was tilting by defining the concept of the slope of the line. In space, we are looking at a plane, and we need to know how it is tilting. To do this, we will multiply two vectors in the plane to get a third vector perpendicular to the plane. The direction of this third vector will tell us the "inclination" of the plane. The product we will use to multiply the vectors together is called the cross product.

Given two vectors, A and B, in space. If A and B are not parallel, then they determine a plane. Now, select a vector n perpendicular to the plane by the right-hand rule. (see figure 1) Remember that n is the unit normal vector.

figure 1

The cross product is A x B = (| A || B | sin q ) n.

There are a few facts that are associated with A x B, and here they are.

FACT:

A x B is orthogonal to both A and B because it is a scalar multiple of n.

FACT:

If one or both of A and B are zero, then A x B = 0.

FACT:

Nonzero vectors A and B are parallel if and only if A x B = 0.

This is the test to determine if two vectors are parallel.

FACT:

Reversing the order of the factors in a nonzero cross product reverses the direction. A x B = - (B x A)

FACT:

i x j = - (j x i) = k

j x k = - (k x j) = i

k x i = - (i x k) = j

i x i = j x j = k x k = 0

FACT:

Because n is a unit vector, the magnitude of A x B is

|A x B| = | A | | B | |sin q | | n |

= | A || B || sin q |.

This is the area of a parallelogram. (see figure 2)

figure 2

Suppose you are turning a bolt with a wrench by applying a force F to the wrench. The torque we produce acts along the axis of the bolt to drive the bolt forward. The magnitude of the torque depends on how far out on the wrench the force is applied and how much of the force is perpendicular to the wrench at the point of application. The number we use to measure the torque's magnitude is the product of the length of the lever are r and the scalar component of F perpendicular to r.

Magnitude of torque vector = | r || F | sin q = | r x F |

If we let n be a unit vector along the axis of the bolt in the direction of torque, then the torque vector is (| r || F | sin q ) n.

How do we find A x B if we do not know the angle between the two vectors A and B? Here is how.

FACT:

Let A = a₁ i + a₂ j + a₃ k and B = b₁ i + b₂ j + b₃ k, then

EXAMPLE 1:

Find the length and direction (when defined) of A x B and B x A when A = 2i - 2j - k and B = i - k.

SOLUTION:

EXAMPLE 2:

Given P (1, 1, 1), Q (2, 1, 3), and R (3, -1, 1), find a) the area of the triangle determined by the points P, Q, and R, and b) the unit vector perpendicular to the plane PQR.

SOLUTION:

First of all, I must construct two vectors out of the three points. I will find the vectors PQ and PR and these two vectors lie in the plane that contains the points P, Q, and R.

PQ = (2 - 1)i + (1 - 1)j + (3 - 1)k = i + 2k

PR = (3 - 1)i + (-1 - 1)j + (1 - 1)k = 2i - 2j

1. If the area of the parallelogram defined by the vectors A and B is

| A x B |, then if I take half of this area, I will have the area of the triangle.

Area of the triangle = (1/2) | A x B |



Area of the triangle = (1/2)(6) = 3

2. The unit vector.

EXAMPLE 3:

Find the area of the parallelogram with vertices A(-6, 0), B(1, -4), C(3, 1), and D(-4, 5).

SOLUTION:

I need to find two vectors. I will find vectors BA and BC. Notice that I do not need the point D.

BA = (-6 - 1)i + (0 - (-4))j = -7i + 4j

BC = (3 - 1)i + (1 - (-4))j = 2i + 5j

Now to find BA x BC.

The area of the parallelogram is | BA x BC |

EXAMPLE 4:

Find the area of the triangle with vertices A (2, 4), B (3, 4), and C (6, 0).

SOLUTION:

BA = (3 - 2)i + (4 - 4)j = -i

BC = (6 - 3)i + (0 - 4)j = 3i - 4j

Area of the triangle ABC = 4/2 = 2.

The product (A x B) · C is called the triple scalar product (or box product) of A, B, and C and | (A x B) · C | = | A x B || C ||cos q | is the volume of the parallelepiped.

FACT:

EXAMPLE 5:

Find (A x B) · C and the volume of the parallelepiped defined by A = i + j - 2k, B = -i - k, and C = 2i + 4j - 2k.

SOLUTION:

Volume of the parallelepiped | A x B | · C = | 8 | = 8