Suppose your are given two vectors A and B, and we place them so their initial points coincide. When the two vectors are placed this way, the vectors form the angle q . (see figure 1) This angle is called the angle between A and B.

figure 1

FACT:

The scalar product (dot product) A · B (A dot B) of vectors A and B is the number A · B = | A || B | cos q where q is the angle between A and B.

FACT:

The dot product is positive when q is acute, and the dot product in negative when q is obtuse.

FACT:

If q = 0, then A makes an angle with itself and A · A = | A || A | cos 0 = | A | ².

FACT:

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

EXAMPLE 1:

Given A = 3i + 2j - k and B = 6i - 4j + 2k, find A · B.

SOLUTION:

A · B = (3)(6) + (2)(-4) + (-1)(2) = 18 - 8 - 2 = 9

One nice outcome from the dot product is the fact that we can determine the angle created by two vectors. We can do this because the definition of the dot product involves the cos q .

FACT:

The angle between two nonzero vectors A and B is

FACT:

Nonzero vectors A and B are perpendicular (orthogonal) if and only if A · B = 0.

This is the test to determine if two vectors are perpendicular to each other.

EXAMPLE 2:

Given A = 2i + 3j and B = i - j + 2k, find (a) A · B, (b) | A |, | B |, (c) the cosine of the angle between A and B, and (d) find q to the nearest hundredth of a radian.

SOLUTION:

A · B = (2)(1) + (3)(-1) + (0)(2) = 2 - 3 = -1

EXAMPLE 3:

Given A = i - j + k and B = 3i + 2j - k, find (a) A · B, (b) | A |, | B |, (c) the cosine of the angle between A and B, and (d) find q to the nearest hundredth of a radian.

SOLUTION:

A · B = (1)(3) + (-1)(2) + (1)(-1) = 3 - 2 - 1 = 0

FACT:

(c A) · B = A · (c B) = c (A · B)

A · ( B + C) = A · B +A · C

The vector projection of B = PQ onto a nonzero vector A = PS is the vector PR determined by dropping a perpendicular from Q to the line PS. (see figure 2)

The notation for this vector is Proj _A B.

figure 2

Why do we want to talk about the projection of one vector onto another vector? In physics, if B represents a force, then Proj _A B represents the effective force in the direction of A. Now let me state the formula for the projection of one vector onto another.

FACT:

FACT:

| B | cos q is called the scalar component of B in the direction of A.

EXAMPLE 4:

Given A = -i + j and B = 2i + j - k, find (a) A · B, | A |, | B |, (b) the cosine of the angle between A and B, (c) the scalar component of B in the direction of A, and (d) the vector projection, Proj _A B.

SOLUTION:

A · B = (-1)(2) + (1)(1) + (0)(-1) = -2 + 1 = -1

EXAMPLE 5:

Find the vector projection, Proj _A B when A = 5j - 3k and B = i + j + k.

SOLUTION:

A · B = (0)(1) + (5)(1) + (-3)(1) = 5 - 3 = 2

A · A = (5)(5) + (-3)(-3) = 25 + 9 = 34

In the study of a topic called mechanics in engineering, there is sometimes a need to express a given vector B as a sum of a vector parallel to A and a vector orthogonal to A.

So B can be expressed as follows.

EXAMPLE 6:

Write B = 2i + j - k as a sum of a vector parallel to A = i + 2j + 3k and a vector orthogonal to A.

SOLUTION:

A · B = (2)(1) + (1)(2) + (-1)(3) = 2 + 2 - 3 = 1

A · A = (1)(1) + (2)(2) + (3)(3) = 1 = 4= 9 = 14

Remember from calculus I that work was defined as W = Fd. This formula holds only if the force is directed along the line of motion. If a force F moving an object through a displacement D = PQ has some other direction, the work is performed by the component of F in the direction of D. If q is the angle between F and D (see figure 3),

then

Work = (scalar component of F in the direction of D)*(length of D) = (| F | cos q ) | D|

= F · D.

EXAMPLE 7:

Find the work done by a force F = 5k (magnitude 5 N) in moving an object along a line from the origin to the point (1, 1, 1). (Distance is in meters.)

SOLUTION:

We need to find the displacement D first. It is the vector from the origin to the point (1, 1, 1).

D = i + j + k

Work = F · D = (0)(1) + (0)(1) + (5)(1) = 5 N - m = 5 J

EXAMPLE 8:

Find the angle between the lines 3x - 2y = 4 and x - 4y = 8.

SOLUTION:

First put each equation into the slope-intercept form.

Now determine the related vectors.

v1 = 2i + 3j

v₂ = 4i + j

v₁ · v₂ = (2)(4) + (3)(1) = 8 + 3 = 11

EXAMPLE 9:

Find the angles between the curves y = x ³ and x = y ². (There will be two intersection points.)

SOLUTION:

First, find the points of intersection of the two curves.

y = (y ²) ³ ® y = y ⁶ ® y ⁶ - y = 0 ® y (y ⁵ - 1) = 0 ® y = 0 or y = 1

When y = 0, then x = 0, and when y = 1, then x = 1.

Now find the derivative of each function.

(0, 0)

m₁ = 0

v₁ is a horizontal vector.

v₁ = i

m₂ is undefined

v₂ is a vertical vector.

v₂ = j

v_{1 ·} v₂ = 0 ® cos q = 0 ® q = p /2

Each point of intersection will have two angles, if q = p /2, then the other angle j = p - p /2 = p /2.

(1, 1)

m₁ = 3

v₁ = i + 3j

m₂ = 1/2

v₂ = 2i + j

v_{1 ·} v₂ = (1)(2) + (3)(1) = 2 + 3 = 5