Algebraic Vectors

ALGEBRAIC VECTORS

Number Pair Notation

To represent a vector numerically, we do this in a very similar manner as coordinate points, i.e. we split the vector movement into a horizontal followed by a vertical component.

Example

a is 2 squares to the right followed by3 squares up.

We write this as

a =

b =

c =

d =

Addition and Subtraction of Numerical Vectors.

Consider followed by

The equivalent is

Note also that + =

When we add or subtract numerical vectors, simply add or subtract the corresponding components:

Example: + =

- =

When we multiply vectors by a number simply multiply each component by that number:

Example:2 x = =

Exercise 3

1. + 2. + 3. +

4. If u = , v = and w = , find (a) u + v(b) u + v + w

5. - 6. - 7. -

8. Find x:

(a) x + = (b) x - =

9. u = and v = , find

(a) 3u + 3v(b) 3(u + v)(c) 5u - 5v(d) 5(u - v)

10. p = and q = , find

(a) 3(p + q)(b) (c) 2(p - 2q) + 3p

Unit Vector Notation

This is another method of writing the components of a vector in an easier form.

We write a unit horizontal movement which in numerical form would be as i.

We write a unit vertical movement which in numerical form would be as j.

Combinations of horizontal and vertical movements are written as combinations if i's and j's.

Example: = 2i + 3j

Multiplying, adding and subtracting vectors in i, j form is the same as that in numerical form - simply combine the appropriate components.

Example:In numerical form + =

In i, j form(3i - 2j) + (4i - j) = 7i - 3j

Exercise 4

1. u = and v = . Find in component form and in terms of i and j

(a) 2u + 5v(b) 4u - 3v

2. u = , v = and w = . Find in component form and in i, j form:

(a) 2v + u(b) 3v - w(c) 2w - u

3. a = , b = and c = . Find in component form and in i, j form:

(a) 3b - c(b) b - 2a(c) 3c - 4b

4. Given that p = 2i - 3j and q = i + 5j, find :

(a) 2p + q(b) p - 3q(c) 3p + 2q

(d) 2(p + q)(e) -4p - q(f) 2p - 4q

Position Vectors

This is the vector from the origin to a given point.

Example:If A is the point (3,4)

OA or a is the position vector of A and a = = 3i + 4j.

Note that the coordinates of A are the components of its position vector.

We can use the position vectors of two points to find the vector joining the two points.

Example:If A is (6,1) and B is (5,-4), find AB.

a = 6i + j

b = 5i - 4j

Looking at this triangle of velocities we have

AB = -a + b = b - a

This rule can be used to find the vector joining any two points if we know the coordinates of the points and hence their position vectors.

In words, the rule states that the vector from one point to another is the position vector of the SECOND point minus the position vector of the FIRST point.

For the above example, AB = b - a

= (6i + j) - (5i - 4j)

= i + 5j or

Example:P is (5,2); Q is (-2, -3). Find PQ.

p = 5i + 2jq = -2i - 3j

PQ = q - p

= (-2i - 3j) - (5i + 2j)

= -7i - 5j

Exercise 5

1. P is the point (2,3) and Q is the point (7,5). Write down the position vectors of OP and OQ in component form. Calculate the components of the vector PQ.

2. Repeat question 1 for the points P(0, -1) and Q (4,2).

3. A is the point (4,0) B is the point (6,2) C is the point (-2, 1).

(a) Find a, b and c.

(b) Find AB, BC and CA.

4. A is (4, 0), B is (-3, 1), C is (0, -6) and D is (1, -3).

For each of these points find its position vector in terms if i and j.

Answers

Exercise 3

1. 2. 3. 4. (a) (b)

5. 6. 7. 8. (a) (b)

9. (a) (b) (c) (d)

10. (a) (b) (c)

Exercise 4

1. (a) , 26i + 11j(b) , -17j

2. (a) , 8j(b) , i + 8j(c) , 6i

3. (a) , -14i + 2j(b) , 6i + 5j(c) , -10i + 7j

4. (a) 5i - j(b) -i - 18j(c) 8i + j(d) 6i + 4j(e) -9i + 7j(f) -26j

Exercise 5

1. OP = , OQ = , PQ =

2. , ,

3. (a) , , (b) , ,

4. a = 4ib = -3i + jc = -6jd = i - 3j

Magnitude or Length of a Vector.

Consider a =

We write the length of a as | a |.We can use Pythagoras's Theorem to find the length of a.

| a | = (32 + 42) = 5

In general, if p = = ai + bj then | p | = (a2 + b2)

Exercise 1

1. Write down the lengths of the following, leaving in square roots where necessary.

(a) (b) (c) (d)

(e) (f)

2. State whether the following is true or false:(a) If |u| = |v| then u = v

(b) If u = v then |u| = |v|

3. Calculate the magnitude of the vector PQ is P and Q are the points:

(a) (5, 0) and (10, 4)(b) (7, 4) and (1, 12)

(c) (-1, -1) and (-5, -6)(d) (4, -1) and (-3, -4)

Answers

Exercise 1

1. (a) 5(b) 65(c) 65(d) 13

2. (a) F(b) T

3. (a) 41(b) 10(c) 41(d) 58

VECTORS IN CARTESIAN AND POLAR FORM

Vectors can be represented mathematically in two ways: Cartesian form (horizontal and vertical components), and polar form (magnitude and direction).

A vector is represented by a straight line whose length is proportional to the magnitude of the vector, and with an arrow pointing in the direction in which the vector is acting.

c

a b

The vector ac can be represented in polar form by the magnitude of ac acting at an angle to the horizontal.

This is written ac = | ac |

It can also be represented in Cartesian (or rectangular) form as a horizontal component ab (or x) and a vertical component bc (or y).

ab = |ac| cos

bc = |ac| sin

For example, a displacement of 5 metres at an angle of 30o to the horizontal can be represented in polar form as 530o

and in Cartesian form as 5 cos 30o = 4.33 horizontally

5 sin 30o = 2.5 vertically.

So the vector 530o can be said to be the resultant of 4.33 horizontally plus 2.5 vertically.

This is written as (4.33, 2.5)

Vectors in Cartesian form can also be converted to polar form as follows:

If the vector ac is represented in Cartesian form as x horizontally and y vertically then

|ac| = (x2 + y2)

and = tan-1 (y/x)

For example, the resultant of a force 6 kN horizontally and a force of 4 kN vertically is a force of

(36 + 16) = 7.21 kN

at an angle of tan-1 (2/3) = 33.82o

This is written as 7.2133.82o

TUTORIAL

Q1.Determine by calculation the magnitude and direction of the force F in each of the following, and hence express in polar form:

a)horizontal component of 6N and a vertical component of 8N.

b)horizontal component of 7N and a vertical component of 3N.

c)x-component of 8.5N and y-component of 6.4N.

d)x-component of 12.3N and y-component of 7.9N.

Q2.Determine by calculation the x and y Cartesian components of the following vectors:

a)8 35ob)2183oc)6.5 54o

d)15.865oe)12.442of)1628o

ANSWERS

Q1.a) 10 53.1ob) 7.623.2oc) 10.6 37o

d) 14.6 32.7o

Q2.a) (6.55, 4.59)b) (2.56, 20.84)c) (3.82, 5.26)

d) (6.68, 14.3) e) (9.21, 8.30)f) (14.13, 7.51)

Dot Product These are vectors:

They can be multiplied using the "Dot Product" CalculatingYou can calculate the Dot Product of two vectors this way:

a b = |a| |b| cos() Where:|a| is the magnitude (length) of vector a|b| is the magnitude (length) of vector b is the angle between a and bSo we multiply the length of a times the length of b, then multiply by the cosine of the angle between a and bOR you can calculate it this way:

a b = ax bx + ay bySo we multiply the x's, multiply the y's, then add.Both methods work!Example: Calculate the dot product of vectors a and b:

a b = |a| |b| cos() a b = 10 13 cos(59.5) a b = 10 13 0.5075...a b = 65.98... = 66 (rounded)a b = ax bx + ay bya b = -6 5 + 8 12a b = -30 + 96a b = 66Both methods came up with the same result (after rounding)Also note that we used minus 6 for ax (it is heading in the negative x-direction)

Right AnglesWhen two vectors are at right angles to each other the dot product is zero.Example: calculate the Dot Product for:

a b = |a| |b| cos() a b = |a| |b| cos(90) a b = |a| |b| 0 a b = 0a b = ax bx + ay bya b = -12 12 + 16 9a b = -144 + 144a b = 0This can be a handy way to find out if two vectors are at right angles.

Three or More DimensionsThis all works fine in 3 (or more) dimensions, too. Example: Find the angle between the following vectors:

We have 3 dimensions, so don't forget the z-components:a b = ax bx + ay by + az bza b = 9 4 + 2 8 + 7 10a b = 36 + 16 + 70 a b = 122Now for the other formula:a b = |a| |b| cos()But what is |a| ? It is the magnitude, or length, of the vector a. We can use Pythagoras: |a| = (42 + 82 + 102) |a| = (16 + 64 + 100) |a| = 180 Likewise for |b|: |b| = (92 + 22 + 72) |b| = (81 + 4 + 49) |b| = 134 And we know from the calculation above that a b = 122, so:a b = |a| |b| cos()122 = 180 134 cos() cos() = 122 / (180 134)cos() = 0.7855... = cos-1(0.7855...) = 38.2...The Dot Product gives a scalar (ordinary number) answer, and is sometimes called the scalar product.

Page VECTOR CROSS PRODUCT

The SCALAR (DOT) PRODUCT is calculated using the following formula:

symbol instead of a DOT.

The magnitude of the resultant vector is given byNote this is similar to the Dot product formula, except sin is used instead of cos.EXAMPLEIf , and , and the angle between the vectors is 40o, find the magnitude of the cross product., so

EXERCISECalculate the CROSS PRODUCT magnitude for the following vectors:1. , , []2. , , []3. , , []4. , , []5. , , []