# MST124 BOOK B UNIT 5

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## 1

Distance formula
The distance between the points (x1,y1) and (x2,y2) is:
sqrt((x2 - x1)2 + (y2 - y1)2

Midpoint formula
The midpoint of the line segment joining the points (x1y1) and (x2,y2) is
((x1+x2)/2) , ((y1+y2)/2))

- if two lines are perpendicular, and not parallel to the axes, then the product of their gradients is -1
- if A and B are points that do not lie on the same horizontal or vertical line, then the gradient of the perpendicular bisector of AB is: -1/gradient of AB

To find the perpendicular bisector of a line segment joining two points (a,b) and (c,d):
- the points (x,y) on this bisector are exactly those that are equidistant from (a,b) and (c,d); that is those satisfying the equation sqrt((x-a)2 + (y-b)2) = sqrt((x-c)2 + (y-d)2)
- so this is the equation of the perpendicular bisector

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## 2

Standard form of the equation of a circle
The circle with centre (a,b) and radius r has equation
(x - a)2 + (y - b)2 = r2

- to find points that lie on a given circle, substitute any particular value of x into the equation of a circle, and solve the resulting equation to find the corresponding values of y
- the equation you have to solve is quadratic, so you obtain two, one, or no values of y
- each of these gives a point on the circle
- you can also sub in a value of y to find a corresponding x value to find (possible) points on a circle

Strategy: To find the equation of the circle passing through three points:
- find the equation of the perpendicular bisector of the line segment that joins any pair of the three points
- find the equation of the perpendicular bisector of the line segment that joins a different pair of the three points
- find the point of intersection of these two lines (the centre of the circle) and the radius (distance from the centre to any of the three points)

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## 3

- a line and a circle can have two, one, or no points of intersection
- to find the points, solve the equation of the line and the equation of the circle simultaneously
- use similar methods to find the points of intersection of a line and a parabola

- two circles may interesect at two points, one point or not at all
- to find the points of intersection, solve their equations simultaneously
- eliminate the terms in x2 and y2 by subtracting one equation from the other

- to specify the positions of points in three-dimensional space, we extend the usual two-dimensional Cartesian coordinate system, by introducing a third axis, the z-axis
- this axis is at right angles to the x- and y-axes

- the two different versions of a three-dimensional coordinate system are known as the right-handed coordinate system and the left-handed coordinate system
- the way of remembering which is which is to use the right-hand rule (figure 22, page 129 book B) and the left-hand rule
- another way to remember is the right-hand grip rule (page 130 book B)

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## 4

Distance formula (three dimensions)
The distance between two points (x1, y1, z1) and (x2, y2, z2) is
sqrt((x2 - x1)2 + (y2 - y1)2 + (z2 - z1)2)

The standard form of the equation of a sphere
The sphere with centre (a,b,c) and radius r has equation
(x - a)2 + (y - b)2 + (z - c)2 = r2

Zero vector
The zero vector, denoted by 0, is the vector whose magnitude is 0 and has no direction

- to vectors are equal if they have the same magnitude and direction

Triangle law for vector addition
To find the sum of two vectors a and b, place the tail of b at the tip of a. Then ab is the vector from the tail of a to the tip of b

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## 5

Parallelogram law for vector addition
To find the sum of two vectors a and b, place their tails together, and complete the resulting figure to form a parallelogram. Then ab is the vector formed by the diagonal of the parallelogram, starting from the point where the tails of a and b meet.

- adding the zero vector to any vector leaves it unchanged
a0a

Negative of a vector
The negative of a vector a, denoted by -a, is the vector with the same magnitude as a, but the opposite direction.

Vector subtraction
To subtract b from a, add -b to a.
aba + (-b

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## 6

Scalar multiple of a vector
Suppose that a is a vector. Then, for any non-zero real number m, the scalar multiple ma of a is the vector:
- whose magnitude is the modulus of m times the magnitude of a
- that has the same direction as a if m is +ve, and the opposite direction is m is -ve
Also, 0a0

Properties of vector algebra
The following properties hold for all vectors ab and c, and all scalars m and n
ab ba
(ab) + ca + (b + c)
a0a
+ (-a) = 0
m(ab) = ma + mb
(m + n)a = ma + na
m(na) = (mn)a
1a

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## 7

- a bearing is an angle between 0o and 360o, measured clockwise in degrees from north to the direction of interest

Opposite, corresponding and alternate angles
Where two lines intersect:
- opposite angles are equal
Where a line intersects parallel lines:
- corresponding angles are equal
- alternate angles are equal
(page 151, Book B)

Watch the video tutorial for Example 13 on page 152 of Book B

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## 8

Component form of a vector
If v = ai + bj, then the expression ai + bj is called the component form of v
- the scalars a and b are called the i-component and the j-component, respectively, of v
If v = ai + bj + ck, then the expression ai + bj + ck is called the component form of
- the scalars a, b and c are called the i-component, j-component and k-component of v

- the i and j components of a two-dimensional vector are alternatively called the x-component and the y-component
- the k-component in a three-dimensional vector can be called the z-component

Alternative component form of a vector
The vector ai + bj can be written in brackets, with a above b
The vector ai + bj + ck can be written in brackets too, with a at the top, b in the middle, and c at the bottom
- a vector written in this form is called a column vector

- the first number in a column vector is its i-component, the second number is its j-component, and, in three dimensions, the third number is its k-component

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## 9

Addition of two-dimensional vectors in component form
If a = a1i + a2j and b = b1+ b2j, then
ab = (a1 + b1)i + (a2 + b2)j

Addition of three-dimensional vectors in component form
if a = a1+ a2j + a3k and b = b1i + b2j + b3k, then
ab = (a1 + b1)i + (a2 + b2)j + (a3 + b3)

Negative of a two-dimensional vector in component form
if b = b1i + b2j, then -b = -b1i - b2

Negative of a three-dimensional vector in component form
if b = b1i + b2j + b3k, then -b = -b1i - b2j - b3k

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## 10

Subtraction of two-dimensional vectors in component form
if a = a1i + a2j and b = b1i + b2j, then
ab = (a1 - b1)i + (a2 - b2)j

Subtraction of three-dimensional vectors in component form
if a = a1i + a2j + a3k and b = b1i + b2j + b3k, then
ab = (a1 - b1)i + (a2 - b2)+ (a3 - b3)

Scalar multiplication of a two-dimensional vector in component form
If = a1i + a2j and m is a scalar, then
ma = ma1i + ma2j

Scalar multiplication of a three-dimensional vector in component form
If a = a1i + a2j + a3k and m is a scalar, then
ma = ma1i + ma2j + ma3k

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## 11

- if P is any point, either in the plane or in three-dimensional space, then the position vector of P is the displacement vector OP, where O is the origin
- the components of the position vector of a point are the same as the coordinates of the point

- if the points A and B have position vectors a and b, respectively then the vector AB is equal to b a

Midpoint formula in terms of position vectors
If the points A and B have position vectors a and b, respectively, then the midpoint of the line segment AB has position vector 1/2(ab

The magnitude of a two-dimensional vector in terms of its components
The magnitude of the vector a+ bj is sqrt(a2 + b2)

The magnitude of a three-dimensional vector in terms of its components
The magnitude of the vector ai + bj + ck is sqrt(a2 + b2 + c2

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## 12

Component form of a two-dimensional vector in terms of its magnitude and its angle with the positive direction
If the two-dimensional vector v makes the angle theta with the +ve x-direction, then
v = magnitude of v*cos(theta)i + magnitude of v*sin(theta)j

- the angle between two vectors a and b is the angle in the range 0 ≤ theta ≤ 180o between their directions when the vectors are placed tail to tail

Scalar product of two vectors
The scalar product of the non-zero vectors a and b is
a.b = magnitude of a * magnitude of b * cos(theta)
where theta is the angle between a and b

- if two non-zero vectors are perpendicular, then their scalar product is zero
- so if the scalar product of two non-zero vectors is 0, then the vectors are perpendicular

- the scalar product of a vector with itself is equal to the square of the magnitude of the vector

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## 13

Properties of the scalar product
The following properties hold for all vectors ab and c, and every scalar m
- suppose that a and b are non-zero. If a and b are perpendicular, then a.b = 0, and vice versa
a.a = (magnitude of a)2
- a.bb.a
a.(b+c) = a.ba.c
- (ma) . b = m(a.b) = a.(mb

Scalar product of two-dimensional vectors in terms of components
If a = a1i + a2j and b = b1i + b2j, then
a.b = a1b1 + a2b2

Scalar product of three-dimensional vectors in terms of components
if a = a1i + a2j + a3k and b = b1i + b2j + b3k then
a.b = a1b+ a2b2 + a3b3

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## 14

Angle between two vectors
The angle theta between any two non-zero vectors a and b is given by
cos(theta) = (a.b)/((magnitude of a)*(magnitude of b))

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