Torque•The magnitude of the torque depends on two things:The magnitude of the torque depends on two things:

•The distance from the axis of the bolt to the point where the force is The distance from the axis of the bolt to the point where the force is applied. This is |applied. This is |rr|, the length of the position vector |, the length of the position vector rr..

•The scalar component of the force The scalar component of the force FF in the direction perpendicular to in the direction perpendicular to rr. This is the . This is the only component that can cause a rotation, and as the figure shows, it is |only component that can cause a rotation, and as the figure shows, it is |FF|sin |sin θθ..

If If nn is a unit vector that points in the direction in is a unit vector that points in the direction in which a right-threaded bolt moves, we define the which a right-threaded bolt moves, we define the torquetorque to be the vector to be the vector ττ = = (|(|rr||||FF|sin |sin θθ) ) nn..

We denote this torque vector by We denote this torque vector by ττ = = rr FF and and we call it the we call it the cross productcross product (or (or vector productvector product) ) of of rr and and FF..

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.

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.

• 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

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| :

Suppose Suppose aa and and bb are given in component form: are given in component form:

Component form

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

a b

ad bcc d

Determinants (cont’d)• For example,

• A determinant of order 3 can be defined in terms of second-order determinants as follows:

2 1

2 4 1 6 146 4

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.

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:

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.

Example

• If 1,3,4 and 2,7, 5 , thena b

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 vectorperpendicular to boththerefore perpendicular to the plane through P, Q, and R. So, lets create vectors…

����������������������������

isPQ PR���������������������������� and and isPQ PR

Solution (cont’d)

• So the vectorto the given plane.

• Any other nonzero scalar multiple of this vector, such asperpendicular to the plane.

40, 15,15 is perpendicular

8, 3,3 , is also

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:

����������������������������

40, 15,15PQ PR

Solution (cont’d)

• The area A of triangle PQR is half the area

of this parallelogram, that is, 5

822

Scalar Triple Product• The product a ∙ (b  c) is called the scalar

triple product of the vectors a, b, and c.

• Its geometricsignificance isillustrated bythe figure.

•The area of the base parallelepiped isThe area of the base parallelepiped isAA = | = |bb      cc|.|.•The height of the parallelepiped is The height of the parallelepiped is hh = | = |aa||cos ||cos θθ|.|.•Thus, the volume of the parallelepiped isThus, the volume of the parallelepiped isVV = = AhAh = | = |bb      cc| || |aa||cos ||cos θθ| = || = |aa  ∙ ( ∙ (bb      cc)|)|

1,4, 7 , 2, 1,4 , and 0, 9,18a b c

Example : show that the vectors are coplanarExample : show that the vectors are coplanar

Solution

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

1,4, 7 , 2, 1,4 , and 0, 9,18a b c

Homework

• P664

-5,7,9,11,15,17,21,23,25