Vectors and Vector Operations



3.1 Dot Products

The dot product of two vectors u and v is a number and it is denoted by u . v. First consider the case where the two vectors u and v are specified by a magnitude and direction. In that case the dot product is defined by

u . v = (magnitude of u) (magnitude of projection of v on u)

= (magnitude of u) (magnitude of v) cos(

= | u | | v | cos(

where

( = angle between u and v

Dot products and work: Suppose an object is acted upon by a constant force F as it moves from point P to point Q. Then the work W done by the force on the object as it moves from P to Q is defined by

W = (length of displacement from P to Q) (

(magnitude of projection of the force in the direction of the displacement)

= . F

Example. Suppose a force of 3 Newtons acts on an object as it undergoes a displacement of 3 meters and the force makes an angle of 60( with the displacement. Find the work done by the force on the object.

Solution.

W = . F = | | | F | cos(

= (3 N) (2 m) cos(60()

= (3)(2)(½) Newton-meters = 3 Joules.

Work is a form of energy and in the metric system the units of energy are Newton-meters or kilogram-meters2 / second2 which are called Joules. The work done by a force on an object shows up in other forms of energy such as the kinetic energy or the potential energy of the object.

For vectors u = and v = that are lists of numbers the dot product is defined as follows.

u . v = u1v1 + u2v2 + u3v3

Proposition. Suppose u and v are vectors that are specified by a magnitude and direction and one introduces a coordinate system so that u and v are represented by lists of numbers, i.e. and v = . Then u . v = u1v1 + u2v2 + u3v3.

Proof. Apply the law of cosines to the triangle whose sides are u and v. One obtains

| v – u |2 = | u |2 + | v |2 – 2 | u | | v | cos(

(v1 – u1)2 + (v2 – u2)2 + (v3 – u3)2 = (u1)2 + (u2)2 + (u3)2 + (v1)2 + (v2)2 + (v3)2 – 2 u . v

-2 (u1v1 + u2v2 + u3v3) = – 2 u . v

u . v = u1v1 + u2v2 + u3v3

Finding the angle between two vectors. We can use the above formulas to find the angle between two vectors that are given in component form, i.e.

( = cos-1

Example. Find the angle between u = 2i – 3j + k and v = i + 2j - k.

Solution. u = and u = , so

( = cos-1 = cos-1 = cos-1

= cos-1 (-0.546) = 2.15 (radians) = 123.1(

Orthogonal vectors. Two vectors u and v are orthogonal if they are perpendicular, i.e. if the angle ( between them is 90(. This occurs precisely if cos( = 0 or u . v = 0.

Example. Let P be the set of vectors u = xi + yj + zk that are orthogonal to the vector v = - i + 2j + 2k. Note that P is a plane passing through the origin. Find the equation that points on this plane satisfy.

Solution. Points on this plane satisfy

u . v = 0

(xi + yj + zk) . (- i + 2j + 2k) = 0

- x + 2y + 2z = 0

Example. Let P be the set of vectors u = xi + yj + zk that satisfy the equation 3x - 2y + z = 0. Describe P geometrically.

Solution.

3x - 2y + z = 0

(xi + yj + zk) . (3i - 2j + k) = 0

u . v = 0

where v = 3i - 2j + k. This is the plane consisting of all vectors u that are orthogonal to v.

Projection of one vector on another. Suppose we are given two vectors u and v and we want to find the projection of v on u. We start with the formula

u . v = (magnitude of u) (magnitude of projection of v on u)

= | u | (magnitude of projection of v on u)

Solving for the magnitude of the projection v on u gives

magnitude of projection of v on u =

In order to get the projection of v on u we have to multiply a unit vector in the direction of u by this. A unit vector in the direction of u is . So

Projection of v on u =

Example. Find the projection of v = 3i + 2j + 2k on u = 2i + 2j + k.

Solution. Projection of v on u = =

= = i + j + k

Algebraic properties of dot products. It is not hard to see that

u . v = v . u (the dot product is symmetric)

u . u = | u |2

(cu) . v = c (u . v) if c is a number (the dot product is homogeneous)

u . (v + w) = u . v + u . w (a distributive law)

i . i = 1 j . j = 1 k . k = 1

i . j = 0 i . k = 0 j . k = 0

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