# Vector Operations

Let’s consider a vector v whose initial point is the * origin* in an xy - coordinate system and whose terminal point is . We say that the vector is in

*and refer to it as a position vector. Note that the ordered pair defines the vector uniquely. Thus we can use to denote the vector. To emphasize that we are thinking of a vector and to avoid the confusion of notation with ordered - pair and interval notation, we generally write*

**standard position**v = < a, b >.

The coordinate a is the *scalar horizontal component* of the vector, and the coordinate b is the

*scalar*of the vector. By

**vertical component****scalar**, we mean a

*numerical*quantity rather than a

*vector*quantity. Thus, is considered to be the

**component form**of v. Note that a and b are NOT vectors and should not be confused with the vector component definition.

Now consider with A = (x_{1}, y_{1}) and C = (x_{2}, y_{2}). Let’s see how to find the position vector equivalent to . As you can see in the figure below, the initial point A is relocated to the origin (0, 0). The coordinates of P are found by subtracting the coordinates of A from the coordinates of C. Thus, P = (x_{2} - x_{1}, y_{2} - y_{1}) and the position vector is .

It can be shown that and have the same magnitude and direction and are therefore equivalent. Thus, = = < x_{2} - x_{1}, y_{2} - y_{1} >.

The * component form* of with A = (x

_{1}, y

_{1}) and C = (x

_{2}, y

_{2}) is

= < x

_{2}- x

_{1}, y

_{2}- y

_{1}>.

**Example 1** Find the component form of if C = (- 4, - 3) and F = (1, 5).

**Solution** We have

= < 1 - (- 4), 5 - (- 3) > = < 5, 8 >.

Note that vector is equivalent to position vector with as shown in the figure above.

Now that we know how to write vectors in component form, let’s restate some definitions.

The length of a vector v is easy to determine when the components of the vector are known. For v = < v_{1}, v_{2} >, we have

|v|^{2} = v^{2}_{1} + v^{2}_{2} **Using the Pythagorean theorem**

|v| = √v^{2}_{1} + v^{2}_{2}.

The * length*, or

*, of a vector v = < v*

**magnitude**_{1}, v

_{2}> is given by |v| = √v

^{2}

_{1}+ v

^{2}

_{2}.

Two vectors are *equivalent* if they have the same magnitude and the same direction.

Let u = < u_{1}, u_{2} > and v = < v_{1}, v_{2} >. Then

< u_{1}, u_{2} > = < v_{1}, v_{2} > if and only if u_{1} = v_{1} and u_{2} = v_{2}.

### Operations on Vectors

To multiply a vector v by a positive real number, we multiply its length by the number. Its direction stays the same. When a vector v is multiplied by 2 for instance, its length is doubled and its direction is not changed. When a vector is multiplied by 1.6, its length is increased by 60% and its direction stays the same. To multiply a vector v by a negative real number, we multiply its length by the number and reverse its direction. When a vector is multiplied by 2, its length is doubled and its direction is reversed. Since real numbers work like scaling factors in vector multiplication, we call them * scalars* and the products kv are called

*of v.*

**scalar multiples**For a real number k and a vector v = < v_{1}, v_{2} >, the * scalar product* of k and v is

kv = k.< v

_{1}, v

_{2}> = < kv

_{1}, kv

_{2}>.

The vector kv is a

*of the vector v.*

**scalar multiple****Example 2** Let u = < - 5, 4 > and w = < 1, - 1 >.Find - 7w, 3u, and - 1w.

**Solution**

- 7w = - 7.< 1, - 1 > = < - 7, 7 >,

3u = 3.< - 5, 4 > = < -15, 12 >,

- 1w = - 1.< 1, - 1 > = < - 1, 1 >.

Now we can add two vectors using components. To add two vectors given in component form, we add the corresponding components. Let u = < u_{1}, u_{2} > and v = < v_{1}, v_{2} >. Then

u + v = < u = < u_{1} + v_{1}, u_{2} + v_{2} >

For example, if v = < - 3, 2 > and w = < 5, - 9 >, then

v + w = < - 3 + 5, 2 + (- 9) > = < 2, - 7 >

If u = < u_{1}, u_{2} > and v = < v_{1}, v_{2} >, then

u + v = < u_{1} + v_{1}, u_{2} + v_{2} >.

Before we define vector subtraction, we need to define - v. The opposite of v = < v_{1}, v_{2} >, shown below, is

- v = (- 1).v = (- 1)< v_{1}, v_{2} > = < - v_{1}, - v_{2} >

Vector subtraction such as u - v involves subtracting corresponding components. We show this by rewriting u - v as u + (- v). If u = < u_{1}, u_{2} > and v = < v_{1}, v_{2} >, then

u - v = u + (- v) = < u_{1}, u_{2} > + < - v_{1}, - v_{2} > = < u_{1} + (- v_{1}), u_{2} + (- v_{2}) > = < u_{1} - v_{1}, u_{2} - v_{2} >

We can illustrate vector subtraction with parallelograms, just as we did vector addition.

### Vector Subtraction

If u = < u_{1}, u_{2} > and v = < v_{1}, v_{2} >, then

u - v = < u_{1} - v_{1}, u_{2} - v_{2} >.

It is interesting to compare the sum of two vectors with the difference of the same two vectors in the same parallelogram. The vectors u + v and u - v are the diagonals of the parallelogram.

**Example 3** Do the following calculations, where u = < 7, 2 > and v = < - 3, 5 >.

a) u + v

b) u - 6v

c)3u + 4v

d)|5v - 2u|

**Solution**

a) u + v = < 7, 2 > + < - 3, 5 > = < 7 + (- 3), 2 + 5 > = < 4, 7 >;

b)u - 6v = < 7, 2 > - 6.< - 3, 5 > = < 7, 2 > - < - 18, 30 > = < 25, - 28 >;

c) 3u + 4v = 3.< 7, 2> + 4.< - 3, 5 > = < 21, 6 > + < - 12, 20 > = < 9, 26 >;

d) |5v - 2u| = |5.< - 3, 5 > - 2.< 7, 2 >| = |< - 15, 25 > - < 14, 4 >| = |< - 29, 21 >| = √(- 29)^{2} + 21^{2} = √1282 ≈ 35,8

Before we state the properties of vector addition and scalar multiplication, we need to define another special vector—the zero vector. The vector whose initial and terminal points are both is the * zero vector*, denoted by O, or < 0, 0 > . Its magnitude is 0. In vector addition, the zero vector is the additive identity vector:

v + O = v. < v

_{1}, v

_{2}> + < 0, 0 > = < v

_{1}, v

_{2}>

Operations on vectors share many of the same properties as operations on real numbers.

*Properties of Vector Addition and Scalar Multiplication*

For all vectors u, v, and w, and for all scalars b and c:

1. u + v = v + u.

2. u + (v + w) = (u + v) + w.

3. v + O = v.

4 1.v = v; 0.v = O.

5. v + (- v) = O.

6. b(cv) = (bc)v.

7. (b + c)v = bv + cv.

8. b(u + v) = bu + bv.

### Unit Vectors

A vector of magnitude, or length, 1 is called a * unit vector*. The vector v = < - 3/5, 4/5 > is a unit vector because

|v| = |< - 3/5, 4/5 >| = √(- 3/5)

^{2}+ (4/5)

^{2}= √9/25 + 16/25 = √25/25 = √1 = 1.

**Example 4** Find a unit vector that has the same direction as the vector w = < - 3, 5 >.

**Solution** We first find the length of w:

|w| = √(- 3)^{2} + 5^{2} = √34. Thus we want a vector whose length is 1/√34 of w and whose direction is the same as vector w. That vector is

u = w/√34 = < - 3, 5 >/√34 = < - 3/√34, 5/√34 >.

The vector u is a unit vector because

|u| = |w/√34| = = √34/34 = √1 = 1.

If v is a vector and v ≠ O, then

(1/|v|)• v, or v/|v|,

is a * unit vector* in the direction of v.

Although unit vectors can have any direction, the unit vectors parallel to the x - and y - axes are particularly useful. They are defined as

i = < 1, 0 > and j = < 0, 1 >.

Any vector can be expressed as a * linear combination* of unit vectors i and j. For example, let v = < v

_{1}, v

_{2}>. Then

v = < v

_{1}, v

_{2}> = < v

_{1}, 0 > + < 0, v

_{2}> = v

_{1}< 1, 0 > + v

_{2}< 0, 1 > = v

_{1}i + v

_{2}j.

**Example 5** Express the vector r = < 2, - 6 > as a linear combination of i and j.

**Solution**

r = < 2, - 6 > = 2i + (- 6)j = 2i - 6j.

**Example 6** Write the vector q = - i + 7j in component form.

**Solution**q = - i + 7j = -1i + 7j = < - 1, 7 >

Vector operations can also be performed when vectors are written as linear combinations of i and j.

**Example 7** If a = 5i - 2j and b = -i + 8j, find 3a - b.

**Solution**

3a - b = 3(5i - 2j) - (- i + 8j) = 15i - 6j + i - 8j = 16i - 14j.

### Direction Angles

The terminal point P of a unit vector in standard position is a point on the unit circle denoted by (cosθ, sinθ). Thus the unit vector can be expressed in component form,

u = < cosθ, sinθ >,

or as a linear combination of the unit vectors i and j,

u = (cosθ)i + (sinθ)j,

where the components of u are functions of the * direction angle* θ measured counterclockwise from the x - axis to the vector. As θ varies from 0 to 2π, the point P traces the circle x

^{2}+ y

^{2}= 1. This takes in all possible directions for unit vectors so the equation u = (cosθ)i + (sinθ)j describes every possible unit vector in the plane.

**Example 8** Calculate and sketch the unit vector u = (cosθ)i + (sinθ)j for θ = 2π/3. Include the unit circle in your sketch.

**Solution**

u = (cos(2π/3))i + (sin(2π/3))j = (- 1/2)i + (√3/2)j

Let v = < v_{1}, v_{2} > with direction angle θ. Using the definition of the tangent function, we can determine the direction angle from the components of v:

**Example 9** Determine the direction angle θ of the vector w = - 4i - 3j.

**Solution** We know that

w = - 4i - 3j = < - 4, - 3 >.

Thus we have

tanθ = (- 3)/(- 4) = 3/4 and θ = tan^{- 1}(3/4).

Since w is in the third quadrant, we know that θ is a third-quadrant angle. The reference angle is

tan^{- 1}(3/4) ≈ 37°, and θ ≈ 180° + 37°, or 217°.

It is convenient for work with applied problems and in subsequent courses, such as calculus, to have a way to express a vector so that both its magnitude and its direction can be determined, or read, easily. Let v be a vector. Then v/|v| is a unit vector in the same direction as v. Thus we have

v/|v| = (cosθ)i + (sinθ)j

v = |v|[(cosθ)i + (sinθ)j] **Multiplying by |v|**

v = |v|(cosθ)i + |v|(sinθ)j.

**Example 10 Airplane Speed and Direction.** An airplane travels on a bearing of 100° at an airspeed of 190 km/h while a wind is blowing 48 km/h from 220°. Find the ground speed of the airplane and the direction of its track, or course, over the ground.

**Solution** We first make a drawing. The wind is represented by and the velocity vector of the airplane by . The resultant velocity vector is v, the sum of the two vectors:

v = + .

The bearing (measured from north) of the airspeed vector is 100°. Its direction angle (measured counterclockwise from the positive x - axis) is 350°. The bearing (measured from north) of the wind vector is 220°. Its direction angle (measured counterclockwise from the positive x - axis) is 50°. The magnitudes of and are 190 and 48, respectively.We have

= 190(cos350°)i + 190(sin350°)j, and

= 48(cos50°)i + 48(sin50°)j.

Thus,

v = +

= [190(cos350°)i + 190(sin350°)j] + [48(cos50°)i + 48(sin50°)j]

= [190(cos350°)i + 48(cos50°)i] + [190(sin350°)j + 48(sin50°)j]

≈ 217,97i + 3,78j.

From this form, we can determine the ground speed and the course:

Ground speed ≈ √(217,97)^{2} + (3,78)^{2} ≈ 218 km/h.

We let α be the direction angle of v. Then

tanα = 3,78/217,97

α = tan^{- 1}3,78/217,97 ≈ 1°.

Thus the course of the airplane (the direction from north) is 90° - 1°, or 89°.

### Angle Between Vectors

When a vector is multiplied by a scalar, the result is a vector. When two vectors are added, the result is also a vector. Thus we might expect the product of two vectors to be a vector as well, but it is not. The * dot product* of two vectors is a real number, or scalar. This product is useful in finding the angle between two vectors and in determining whether two vectors are perpendicular.

The **dot product** of two vectors u = < u_{1}, u_{2} > and v = < v_{1}, v_{2} > is

u • v = u_{1}.v_{1} + u_{2}.v_{2}

(Note that u_{1}v_{1} + u_{2}v_{2} is a *scalar*, not a vector.)

**Example 11** Find the indicated dot product when

u = < 2, - 5 >, v = < 0, 4 > and w = < - 3, 1 >.

a)u • w

b)w • v

**Solution**

a) u • w = 2(- 3) + (- 5)1 = - 6 - 5 = - 11;

b) w • v = (- 3)0 + 1(4) = 0 + 4 = 4.

The dot product can be used to find the angle between two vectors. The angle *between* two vectors is the smallest positive angle formed by the two directed line segments. Thus the angle θ between u and v is the same angle as between v and u,and 0 ≤ θ ≤ π.

If θ is the angle between two nonzero vectors u and v, then

cosθ = (u • v)/|u||v|.

**Example 12** Find the angle between u = < 3, 7 > and v = < - 4, 2 >.

**Solution** We begin by finding u • v, |u|, and |v|:

u • v = 3(- 4) + 7(2) = 2,

|u| = √3^{2} + 7^{2} = √58, and

|v| = √(- 4)^{2} + 2^{2} = √20.

Then

cosα = (u • v)/|u||v| = 2/√58.√20

α = cos^{- 1}(2/√58.√20)

α ≈ 86,6°.

### Forces in Equilibrium

When several forces act through the same point on an object, their vector sum must be O in order for a balance to occur. When a balance occurs, then the object is either stationary or moving in a straight line without acceleration. The fact that the vector sum must be O for a balance, and vice versa, allows us to solve many applied problems involving forces.

**Example 13 Suspended Block.** A 350-lb block is suspended by two cables, as shown at left. At point A, there are three forces acting: W, the block pulling down, and R and S, the two cables pulling upward and outward. Find the tension in each cable.

**Solution** We draw a force diagram with the initial points of each vector at the origin. For there to be a balance, the vector sum must be the vector O:

R + S + W = O.

We can express each vector in terms of its magnitude and its direction angle:

R = |R|[(cos125°)i + (sin125°)j],

S = |S|[(cos37°)i + (sin37°)j], and

W = |W|[(cos270°)i + (sin270°)j]

= 350(cos270°)i + 350(sin270°)j

= -350j cos270° = 0; sin270° = - 1.

Substituting for R, S, and W in R + S + W + O, we have

[|R|(cos125°) + |S|(cos37°)]i + [|R|(sin125°) + |S|(sin37°) - 350]j = 0i + 0j.

This gives us a system of equations:

|R|(cos125°) + |S|(cos37°) = 0,

|R|(sin125°) + |S|(sin37°) - 350 = 0.

Solving this system, we get

|R| ≈ 280 and |S| ≈ 201.

The tensions in the cables are 280 lb and 201 lb.