2. Torque
The magnitude of the torque depends on two things:
The distance from the axis of the bolt to the point where the force is applied. This is |r|, the length of the position vector r.
The scalar component of the force F in the direction perpendicular to r. This is the only component that can cause a rotation, and as the figure shows, it is |F|sin θ.
If n is a unit vector that points in the direction in which a right-threaded bolt moves, we define the torque to be the vector τ = (|r||F|sin θ) n.
We denote this torque vector by τ = r  F and we call it the cross product (or vector product) of r and F.

4. Example
A bolt is tightened by applying a 40-N force to a 0.25-m wrench, as shown.
Find the magnitude of the torque about the center of the bolt.

5. Solution The magnitude of the torque vector is
If the bolt is right-threaded, then the torque vector itself is
τ = |τ| n ≈ 9.66n
where n is a unit vector directed down into the screen.

6. Find i  j and j  i.
The standard basis vectors i and j both have length 1 and the angle between them is π/2.
By the right-hand rule, the unit vector perpendicular to i and j is n = k
i  j = ((|i| |j|) sin(π/2)) k = k
But if we apply the right-hand rule to the vectors j and i (in that order), we see that n points downward and so n = –k.
j  i = –k

7. Geometric Interpretation
The vectors a and b determine a parallelogram with base |a|, altitude |b|sin θ, and area
A = |a|(|b|sin θ) = |a  b| :

8. Component form
Suppose a and b are given in component form:

9. Determinants
In order to make this expression for a  b easier to remember, we use the notation of determinants.
A determinant of order 2 is defined by

10. Determinants (cont’d)
For example,
A determinant of order 3 can be defined in terms of second-order determinants as follows:

11. Determinants (cont’d)
For example,
Now we rewrite our earlier formula for a  b using second-order determinants and the standard basis vectors i, j, and k.

12. Determinants (cont’d)
The cross product of a = a1 i + a2 j + a3 k and b = b1 i + b2 j + b3 k is
Using a third-order determinant we can condense this formula even further:

13. Determinants (cont’d)
Although the first row of the symbolic determinant on the right side consists of vectors, we can expand it as though it were an ordinary determinant.

14. Example If

15. Example
Find a vector perpendicular to the plane that passes through the points P(1, 4, 6), Q(–2, 5, –1), and R(1, –1, 1).
Solution The vector perpendicular to both therefore perpendicular to the plane through P, Q, and R. So, lets create vectors…

16. Solution (cont’d) So the vector to the given plane.
Any other nonzero scalar multiple of this vector, such as perpendicular to the plane.

18. Example
Find the area of the triangle with vertices P(1, 4, 6), Q(–2, 5, –1), and R(1, –1, 1).
Solution In the preceding example we showed that
The area of the parallelogram with adjacent sides PQ and PR is the length of this cross product:

19. Solution (cont’d)
The area A of triangle PQR is half the area of this parallelogram, that is,

20. Scalar Triple Product
The product a ∙ (b  c) is called the scalar triple product of the vectors a, b, and c.
Its geometric significance is illustrated by the figure.
The area of the base parallelepiped is A = |b  c|.
The height of the parallelepiped is h = |a||cos θ|.
Thus, the volume of the parallelepiped is
V = Ah = |b  c| |a||cos θ| = |a ∙ (b  c)|

21. Example : show that the vectors are coplanar

22. Solution
The volume of the parallelepiped determined by a, b, and c is 0, so these vectors are coplanar.

23. Homework
P664
-5,7,9,11,15,17,21,23,25