Spatial Sense

In this section, you will learn how to derive the circle equation. You will also learn how to manipulate this equation under given conditions and apply it to solve real-life problems. The content of this section is imperative as it forms the foundation on which other advanced topics in mathematics are built. Its real-life applications are vast so why not delve further through your own research.

KEY IDEAS

• A circle is a path traced by all points in a plane which are equidistant from a fixed point in the plane. The fixed point is called the centre.

• The standard equation of a circle with centre (h, k) and radius r is given as:

• (x − h)²+ (y − k)²= r²• The general equation of a circle with centre ( − g, − f) and radius r is given as:

• x²+ y²+ 2gx + 2fy + c = 0

A Circle The circle is the path traced by all points in a plane which are equidistant from a fixed point in the plane. The fixed point is called the centre.

Let us use the activity below to revise the parts of the circle. You will need a graph sheet and a set of mathematical instruments to do this activity.

Activity 4.1- Parts of a circle

Step 1: Make a copy of the graph below.

Figure 4.1: A circle

Step 2: Calculate |OA|, |OB| and |OC|

Step 3: Comment on any observation from your result in Step 2

Step 4: Aside from points A, B and C, list the coordinates of any two points on the circle’s circumference. Find the distance between these points and point O. Comment on your observations.

Step 5: Join points O and C with a straight line.

Measure the angle between OC and the x− axis.

Let us use our diagram to revise the parts of the circle.

Lines OA, OB and OC are the radii of the circle. From the activity, |OA|=|OB|=|OC|.

Line AB is an example of a chord. Any straight line that connects any two points on the circumference of the circle is a chord.

Figure 4.2: A circle The diameter is an example of a chord. It is the longest chord. The diameter is a straight line from a point on the circumference through the centre to another point on the circumference of the circle. In the above diagram, line COD is a diameter.

The x− axis is tangent to the circle at point C(1, 0). The tangent touches the circle at one point. The angle between the tangent and the radius or the diameter at the point of tangency is 90° . In the diagram OCX = 90° The area, shaded red, between the arc AB and the line AB is called a segment.

Figure 4.3: A segment The area bounded by the arc AB and the lines OA and OB is called a sector.

Figure 4.4: A sector

Example 4.1

The diagram below shows the graph of a circle. Use it to answer the questions.

Figure 4.5: Graph of a circle

a. State the coordinates of the centre.

b. Find the radius of the circle

c. Find the length of chord AB

Solution

a. Coordinate of the centre is (10, -5)

b. The radius of the circle is 10 units

c. The coordinate of point A is (10, -15) and that of B is (20, -5)

d. |AB|= √_________________ (20 − 10)²+ ( − 5—15 )²e. |AB|= √1 0²+ 10²f. = √100 + 100

g. = √200

h. = √100 × 2

i. = √100 × √2 j. = 10 √2 units.

The equation of a circle gives information about the centre and radius of the circle.

Figure 4.6: A circle The equation of a circle with centre (h, k) and radius r is given as:

(x − h)²+ (y − k)²= r², where (x, y) is any point on the circumference of the circle.

How did we come by this formula? The following activity will demonstrate.

Activity 4.2: Standard Equation of a Circle

Step 1: Make a copy of the graph below. (a, b) is the centre of the circle and (x, y) is any point on the circumference of the circle.

Figure 4.7: Graph of a circle

Step 2: Using the radius as the hypotenuse, draw a right-angle triangle. Find its horizontal and vertical distance.

Step 3: Using the Pythagoras theorem, write an equation to connect the lengths of sides of the triangle.

Your answer should be similar to the one below:

Figure 4.8: Graph of a circle From the figure, we have a right-angle triangle with hypotenuse, r, and the lengths of the other sides as (x − h) and (y − k) Applying Pythagoras theorem, this gives:

(x − h )²+ (y − k )²= r²Alternatively, you can find the distance between the centre (h, k) and any point on the circumference (x, y) and you will achieve the same result r = √____________ (x − h)²+ (y − k )²Square both sides of the equation:

(r )²= [√____________ (x − h)²+ (y − k )²]²r²= (x − h )²+ (y − k )²Rearranging it, we have:

(x − h )²+ (y − k )²= r²This gives the standard equation of a circle.

Let us modify this equation for a few special scenarios.

Scenario 1: When the centre of the circle is the origin (0, 0).

Figure 4.9: Graph of a circle When (h, k) = (0, 0), the equation simplifies to:

(x − 0)²+ (y − 0)²= r²x²+ y²= r² Scenario 2: When the centre is any point on the x-axis other than (0, 0).

Remember that on the x-axis, y = 0.

Figure 4.10: Graph of a circle So, we will use (h, k) = (h, 0) (x − h)²+ (y − 0)²= r²(x − h)²+ y²= r²Scenario 3: When the centre is any point on the y-axis other than (o, 0).

Remember that on the y-axis, x=0.

Figure 4.11: Graph of a circle So, we will use (h, k) = (0, k) (x − 0)²+ (y − k)²= r²x²+ (y − k)²= r²

Example 4. 2

Write the standard form of the circle equation with the following properties:

a. Centre (1, 7) and radius 5

b. Centre (3, 0) and radius 2/3

c. Centre (0, -5) and radius 2

d. Centre (-6, -7) and radius 1

Solution

The standard equation of a circle is given by (x − h)²+ (y − k)²= r²a. Centre (1, 7) and radius 5 Using the standard equation, replace h with 1, k with 7 and r with 5 (x − 1)²+ (y − 7)²= 5²(x − 1)²+ (y − 7)²= 25

b. Centre (3, 0) and radius 2/3

c. using the standard equation, replace h with 3, k with 0 and r with 2/3

d. (x − 3)²+ (y − 0)²= (2/3) 2

e. (x − 3)²+ y²= 4/9

f. Centre (0, -5) and radius 2

g. (x − 0)²+ (y − ( − 5))²= 2²h. x²+ (y + 5)²= 4

i. Centre (-6, -7) and radius 1 (x − ( − 6))²+ (y − ( − 7))²= 1²(x + 6)²+ (y + 7)²= 1

Example 4.3

1. The diagram shows the graph of circles A, B and C. Use them to answer the following questions.

Figure 4.12: Graph of circles A, B and C

a. Find the diameter of circle:

i. A

ii. B

iii. C

b. Write in standard form the equation of:

i. A

ii. B

iii. C

2. A phone’s WI-FI has a range of 10m in all directions. If the phone is 2m West and 5m North, write an equation to represent the area within which the phone’s Wi-Fi can operate.

Solution

1. a.

i. Diameter of A= 4 units

ii. Diameter of B= 6 units

iii. Diameter of C= 3 units b.

i. The coordinates of the centre of circle A is ( − 2, 2).

Since the diameter is 4 units, it means the radius = 1/2 × 4 = 2 So, the equation of circle A is: (x − ( − 2))²+ (y − 2)²= 2²(x + 2)²+ (y − 2)²= 4

ii. The coordinates of the centre of circle B is (2, 0).

Since the diameter is 6 units, it means the radius = 1/2 × 6 = 3 So, the equation of circle B is: (x − 2)²+ (y − 0)²= 3²(x − 2)²+ y²= 9

iii. The coordinates of the centre of circle C is ( − 2, − 3).

Since the diameter is 3 units, it means the radius = 1/2 × 3 = 3/2 So, the equation of circle C is: (x − ( − 2))²+ (y − ( − 3))²= (3/2) 2 (x + 2)²+ (y + 3)²= 9/4 2.

Figure 4.13: Graph of the range of the wi-fi signal The range of the wi-fi signal is in the form of a circle with a radius of 10.

2 units west and 5 units north are the same as the coordinates (− 2, 5). This point gives the centre of the wi-fi signal.

The equation of the circle representing the range of the phone’s wi-fi signals is:

(x − ( − 2))²+ (y − 5)²= 10²(x + 2)²+ (y − 5)²= 100

Example 4.4

A Trotro driver operates within 40km at all angles from a lorry station. The lorry station is located 15km south and 8km east of the driver’s house.

a. Write an equation to represent the driver’s travel boundary from the lorry station using his house as the origin.

b. Find the furthest distance between the driver’s house and his travel boundary.

c. Find the shortest distance between the driver’s house and his travel boundary.

Solution

Figure 4.14: Graph of the lorry station and the driver’s house

a. The driver’s travel boundary is in the form of a circle with a radius of 40km.

The centre is the lorry station. From his house, the lorry station is 15km south which represents y = − 15 and 8km east which represents x = 8

b. Thus, the centre = (8, − 15) The equation of the driver’s travel boundary is:

(x − 8)²+ (y − ( − 15))²= 40²(x − 8)²+ (y + 15)²= 40²

c. We first find the distance between the driver’s house and the lorry station.

Distance = √__________ 8²+ ( − 15 )²= √64 + 225 = √289 = 17 km Since his travel distance is 40km, it means the furthest the driver travels from his house is 17km +40km = 57km The shortest distance between the driver’s house and his travel boundary is 40 − 17 = 23 km The general equation of a circle The general equation of a circle is: x²+ y²+ 2gx + 2fy + c = 0 To derive this formula, we will modify the standard equation.

We are going to replace the centre (h, k) with ( − g, − f).

Recall that the standard equation of a circle is (x − h)²+ (y − k)²= r²where (h, k) is the centre and r is the radius.

Replacing (h, k) with (− g, − f) we have:

(x − (− g))²+ (y − (− f))²= r²(x + g)²+ (y + f)²= r²Expanding the resulting equation, we get:

(x + g)(x + g) + (y + f)(y + f) = r²x²+ gx + gx + g²+ y²+ fy + fy + f²= r²x²+ 2gx + g²+ y²+ 2fy + f²= r²x²+ y²+ 2gx + 2fy + g²+ f²− r²= 0 Since g²+ f²− r²will simplify into a constant, replace it with C This gives, x²+ y²+ 2gx + 2fy + c = 0

Note that when the equation is written in the general form, the centre = (− g, − f) = ( 2g___ − 2, 2f___ − 2) = (Coefficient of x___________

–2 , Coefficient of y___________ − 2 ) = − 1/2( Coefficient of x, Coefficient of y) Also, C = g²+ f²− r². Solving for r, we have r²= g²+ f²− c √r²= √g²+ f²− c r = √g²+ f²− c

Example 4.5

Write the general equation of a circle with:

a. Centre (1, 3) and radius 2.

b. Centre ( − 2, 4) and radius 1/2

Solution

Start from the standard equation and work through it to get the general equation.

a. Centre (1, 3) and radius 2.

(x − 1)²+ (y − 3)²= 2²(x − 1)(x − 1) + (y − 3)(y − 3) = 4 x²− 2x + 1 + y²− 6y + 9 = 4 x²+ y²− 2x − 6y + 9 + 1 − 4 = 0 x²+ y²− 2x − 6y + 6 = 0 Alternatively, we can solve for the variables in the general equation and then substitute the answer into the general equation.

x²+ y²+ 2gx + 2fy + c = 0 The centre is = (− g, − f) = (1, 3) ⟹ g = − 1 and f = − 3 C = (− 1)²+ (− 3)²− 2²C = 1 + 9 − 4 = 6 Substituting these values into the general equation gives:

x²+ y²+ 2( − 1)x + 2( − 3)y + 6 = 0 x²+ y²− 2x − 6y + 6 = 0

b. Centre (− 2, 4) and radius 1/2 (x − ( − 2))²+ (y − 4)²= (1/2) 2 (x + 2)(x + 2) + (y − 4)(y − 4) = 1/4 x²+ 4x + 4 + y²− 8y + 16 = 1/4 x²+ y²+ 4x − 8y + 20 − 1/4 = 0 x²+ y²+ 4x − 8y + 79/4 = 0 Multiply through by 4, so we only have integer values:

4 x²+ 4 y²+ 16x − 32y + 79 = 0 Alternatively, we can solve for the variables in the general equation and then substitute the answer into the general equation.

x²+ y²+ 2gx + 2fy + c = 0 The centre is (− g, − f) = (− 2, 4) ⟹ g = 2 and f = − 4 C = (2)²+ (− 4)²− (1_ 2) 2 C = 4 + 16 − 1/4 C = 20 − 1/4 = 79/4 Substituting these values into the general equation gives:

x²+ y²+ 2(2)x + 2( − 4)y + 79/4 = 0 x²+ y²+ 4x − 8y + 79/4 = 0 Multiply through by 4 4 x²+ 4 y²+ 16x − 32y + 79 = 0

Example 4.6

Find the centre and radius of the following circle equations

a. x²+ y²− 4x − 2y − 4 = 0

b. x²+ y²− 8x + 6y = 0

c. x²+ y²+ 6x − 40 = 0

d. 4 x²+ 4 y²− 8x + 3 = 0

Solution

a. x²+ y²− 4x − 2y − 4 = 0 Method 1: Using completing the square method x²− 4x + y²− 2y = 4 group corresponding terms ( x²− 2x − 2x + 4) + ( y²− y − y + 1) = 4 + 4 + 1 Expand and complete the square (x − 2 )²+ (y − 1 )²= 9 (x − 2 )²+ (y − 1 )²= 3²This shows that the centre is (2, 1) and the radius is 3 Method 2: Alternatively, compare the given equation to the general equation of the circle and solve for the centre and the radius.

The general equation of a circle is x²+ y²+ 2gx + 2fy + c = 0 The given equation is x²+ y²− 4x − 2y − 4 = 0 x²+ y²+ 2(− 2)x + 2(− 1)y + ( − 4) = 0 By comparing coefficients, we have g = − 2 ⟹ − g = 2 Also, f = − 1 ⟹ − f = 1 and c = − 4 Recall that when the circle equation is written in the general form, the centre is ( − g, − f) Therefore, the centre is (2, 1).

You will achieve the same result if you divide the coefficient of x and y by − 2 Centre = (− 4 _ − 2 , − 2 _ − 2) = (2, 1) Radius = r = √g²+ f²− c r = √__________ 2²+ 1²− ( − 4) = √9 = 3

b. x²+ y²− 8x + 6y = 0 Method 1: Using the completing the square method, we have:

x²− 8x + 16 + y²+ 6y + 9 = 16 + 9 (x − 4 )²+ (y + 3 )²= 25 (x − 4 )²+ (y + 3 )²= 5²Therefore, the centre is (4, -3) and the radius is 5 Method 2: Alternatively, we can re-write the equation as x²+ y²+ 2(− 4)x + 2(3)y + 0 = 0 Comparing this equation to x²+ y²+ 2gx + 2fy + c = 0 , we have:

g = − 4 ⟹ − g = 4 , f = 3 ⟹ − f = − 3 and C=0 Therefore, the centre is (4, -3) Radius = r = √____________ (− 4)²+ 3²− 0 = √25 = 5

c. x²+ y²+ 6x − 40 = 0 Method 1: Completing the square, we have:

x²+ 6x + 9 + y²+ 2(0)x + 0 = 40 + 9 + 0 (x + 3 )²+ (y + 0 )²= 49 (x + 3 )²+ (y + 0 )²= 7²Centre is (-3, 0) and radius is 7 Method 2: Alternatively, we can re-write the equation as x²+ y²+ 2(3)x + 2(0)y − 40 = 0 Comparing this equation to x²+ y²+ 2gx + 2fy + c = 0 , we have:

g = 3 ⟹ − g = − 3 , f = 0 ⟹ − f = 0 and C= − 40 Therefore, the centre is (-3, 0) Radius = r = √___________ (3)²+ 0²− (− 40) = √49 = 7

d. 4 x²+ 4 y²− 8x + 3 = 0 Method 1: Completing the square, we will first divide the equation by 4 x²+ y²− 2x + 3/4 = 0 x²− 2x + 1 + y²+ 0 = −3/4 + 1 + 0 (x − 1 )²+ (y + 0 )²= 1 _ 4 (x − 1 )²+ (y + 0 )²= (1_ 2) 2 Centre is (1, 0) and radius is 1/2 Method 2: Alternatively, we can re-write the equation as:

x²+ y²+ 2(− 1)x + 2(0)y + 3/4 = 0 Comparing this equation to x²+ y²+ 2gx + 2fy + c = 0 , we have:

g = − 1 ⟹ − g = 1 , f = 0 ⟹ − f = 0 and C= 3/4 Therefore, the centre is (1, 0) Radius = r = √____________ (− 1)²+ 0²− 3/4 = √_ 1/4 = 1/2

How to Find the Equation of the Circle Given the Endpoints of the Diameter Method 1:

Step 1: Find the centre of the circle by finding the mid-point of the diameter.

Step 2: Calculate the length of the radius. This is half the diameter and is the same as the distance between the centre and any of the endpoints of the diameter.

Step 3: Substitute the centre and the radius into the standard equation of the circle.

Step 4: Simplify the resulting equation to achieve the desired result.

Method 2:

An alternative method involves using the relationship between gradients of two perpendicular lines.

Recall that if m₁ and m₂ are gradients of any two lines which intersect at right angles, then m₁ × m₂ = − 1 Also, if one side of a triangle inscribed in a circle is a diameter, then the triangle is right-angled. The angle opposite the diameter is the right angle. This property of circles is illustrated in the diagram below.

Figure 4.15: Graph of circle with centre O From the figure, A( x₁, y₁) and B( x₂, y₂) are the endpoints of the diameter and P(x, y) is any point on the circumference of the circle. The angle opposite the diameter is AOB= 90° . This means that the product to the gradients of lines AP and BP is equal to − 1 . Using this relation will generate the equation of the circle.

Gradient ( m₁) of |AP| = y − y₁_____ x − x₁ Gradient (m₂) of |BP| = y − y₂_____ x − x₂ Using m₁ × m₂ = − 1 , we have y − y₁_ x − x₁ × y− y₂_____ x − x₂ = − 1

Example 4.7

The endpoints of the diameter of a circle are (-2, 2) and (4, -6).

Find the;

a. radius and centre of the circle

b. general equation of the circle.

Solution

The graph below shows a sketch of the circle.

Figure 4.16: Graph of circle with diameter (-2, 2) and (4, -6) Method 1:

1. The centre is equal to the mid-point of the endpoints of the diameter

2. Midpoint = 1/2(− 2 + 4, 2 + (− 6))

3. = 1/2(2, − 4)

4. = (2/2, −4/2) = (1, − 2) Radius is the distance between the centre and any of the endpoint/point on the circumference.

r = √_______________ (− 6 − (− 2))²+ (4 − 1 )²r = √(− 4)²+ 3²r = √16 + 9 = √25 = 5 Using (x − h )²+ (y − k )²= r², we have:

(x − 1 )²+ (y − (− 2) )²= 5²(x − 1 )²+ (y + 2 )²= 25 x²− 2x + 1 + y²+ 4y + 4 = 25 x²+ y²− 2x + 4y + 5 = 25 x²+ y²− 2x + 4y + 5 − 25 = 0 x²+ y²− 2x + 4y − 20 = 0 Method 2:

Alternatively, let any point on the circumference of the circle be P(x, y) as shown in the figure below.

Figure 4.17: Graph of circle with diameter (-2, 2) and (4, -6) The gradient of the line connecting (-2, 2) and (x, y) = − 2/x − (− 2) = y − 2/x + 2 The gradient of the line connecting (4, -6) and (x, y) = y − ( − 6)_______ x − 4 = y + 6/x − 4 Since the two lines intersect at 90° , (y − 2_ x + 2) × (y + 6_ x − 4) = − 1 (y − 2)(y + 6)_ (x + 2)(x − 4) = − 1 y²+ 6y − 2y − 12/x²− 4x + 2x − 8 = − 1 y²+ 4y − 12_ x²− 2x − 8 = −1_ 1 Cross multiply:

y²+ 4y − 12 = − 1(x²− 2x − 8) y²+ 4y − 12 = −x²+ 2x + 8 Regroup the terms:

y²+ x²− 2x + 4y − 12 − 8 = 0 y²+ x²− 2x + 4y − 20 = 0 Equation of a Circle Given Three Points on the Circumference of the Circle Method 1:

Substitute each of the points into the general equation of the circle (x²+ y²+ 2gx + 2fy + c = 0).

This results in 3 simultaneous equations with 3 unknowns to be solved for g, f and c .

Method 2:

Alternatively, follow the following steps.

Step 1: Find the equation of the perpendicular bisectors of any two of the chords.

In the Figure 4.18 A, B and C are points on the circumference of the circle.

Figure 4.18: Graph of circle with equation (x²+ y²+ 2gx + 2fy + c = 0)

Step 2: Solve the equations obtained in step 1 simultaneously. This will give the coordinates of the centre of the circle.

Step 3: Using the centre and any point on the circumference, calculate the radius.

Step 4: Use the radius and the centre to calculate the equation of the circle.

Example 4.8

Find the equation of the circle which satisfies the points (2, -4), (-6, 4) and (2, 12).

Solution

Method 1:

In the diagram below, A, B and C are points on the circumference of the circle.

The perpendicular bisectors of chords AB and BC intersect at point V. Point V is the centre of the circle.

Figure 4.19: Graph of circle with points (2, -4), (-6, 4) and (2, 12) Substitute the points into the general equation of the circle.

Substituting (-6, 4) into x²+ y²+ 2gx + 2fy + c = 0 gives:

( − 6)²+ 4²+ 2g( − 6) + 2f(4) + c = 0 36 + 16 − 12g + 8f + c = 0 − 12g + 8f + c = − 52 ……………………..(1) Substituting (2, 12) into the same equation:

(2)²+ (12)²+ 2g(2) + 2f(12) + c = 0 4 + 144 + 4g + 24f + c = 0 4g + 24f + c = − 148 ………………………...(2) Substituting (2, -4) into the same equation:

(2)²+ (− 4)²+ 2g(2) + 2f(− 4) + c = 0 4 + 16 + 4g − 8f + c = 0 4g − 8f + c = − 20 ……………………………. (3) Subtract (2) from (3). That is (3) – (2) (4g − 8f + c) − (4g + 24f + c) = − 20 − ( − 148) 4g − 8f + c − 4g − 24f − c = − 20 + 148 − 32f = 128 f = − 128/32 = − 4 Subtract (1) from (2):

(4g + 24f + c) − (− 12g + 8f + c) = − 148 − (− 52) 4g + 24f + c + 12g − 8f − c = − 148 + 52 16g + 16f = − 96 16g + 16(− 4) = − 96 16g − 64 = − 96 16g = − 96 + 64 16g_ 16 = − 32/16 g = − 2 Substitute g = − 2 and f = − 4 into (3):

4(− 2) − 8(− 4) + c = − 20 − 8 + 32 + c = − 20 c = − 20 + 8 − 32 c = − 44 Finally, substitute g , f and c into the general equation.

x²+ y²+ 2( − 2)x + 2( − 4)y − 44 = 0 x²+ y²− 4x − 8y − 44 = 0 gives the equation of the circle which satisfies (2, − 4), ( − 6, 4) and (2, 12).

Method 2:

First plot the points on the x-y plane and connect the points with chords.

Figure 4.20: Graph of the x-y plane Mid-point of ( − 6, 4) and (2, 12) is (− 6 + 2/2 , 4 + 12/2 ) = ( − 2, 8) The gradient of the line connecting the points ( − 6, 4) and (2, 12) is 12 − 4/2 − (− 6) = 8/8 = 1 The gradient of the perpendicular bisector is − 1 The equation of the perpendicular bisector is y − 8 = − 1(x − ( − 2)) y = − 1(x + 2) + 8 y = − x − 2 + 8 y = − x + 6 …………………….(1) Mid-point of ( − 6, 4) and (2, − 4) is (− 6 + 2/2 , 4 − 4/2 ) = ( − 2, 0) The gradient of the line connecting the points ( − 6, 4) and (2, − 4) is − 4 − 4/2 − (− 6) = − 8/8 = − 1 The gradient of the perpendicular bisector is 1 The equation of the perpendicular bisector is y − 0 = 1(x − ( − 2)) y = (x + 2) y = x + 2 y = x + 2 …………………….(2) Add equation 1 to equation 2:

y + y = − x + x + 6 + 2 2y = 8 y = 4 Substitute y = 4 into equation 2 and solve for x :

4 = x + 2 x = 4 − 2 x = 2 This means the coordinates of the centre of the circle is (2, 4).

The distance between the centre and any point on the circle’s circumference gives the radius.

Distance between (2, 4) and (2, 12)= r = √_____________ (2 − 2)²+ (12 − 4)²= √0²+ 8²= √64 = 8 Finally, substitute the centre, (2, 4), and the radius, 8, into the standard equation of the circle.

(x − 2)²+ (y − 4)²= 8²x²− 4x + 4 + y²− 8y + 16 = 64 x²+ y²− 4x − 8y + 20 − 64 = 0 x²+ y²− 4x − 8y − 44 = 0 A construction of the circle is shown below

Figure 4.21: Graph of circle x²+ y²− 4x − 8y − 44 = 0

Study Figure 4.22 carefully:

Figure 4.22: Tangent of a circle In Figure 4.22, V is the circle’s centre, |VO| is a radius and AOB is a tangent to the circle at O. The tangent is a line that touches the circle at only one point.

The equation of the radius, VO, or its extension is the normal to tangent, AOB.

The normal and the tangent intersect at right angles at the point of tangency.

Since |VO| and |AOB| intersect at right angles, We have the (gradient of |OV| )× (gradient of |AOB|) = − 1 Let us use this relation to answer a few questions.

How to find the equation of a tangent and normal given the circle equation and a point, (a, b), on the circumference of the circle

Step 1: Using the given circle equation, find the centre, (h, k)

Step 2: Find the gradient of the normal. This is the gradient (m) of the line connecting points (h, k) and (a, b)

Step 3: Find the gradient of the tangent. The gradient of the tangent is − 1/m, where m is the gradient of the normal.

Step 4: Find the equation of the tangent. This is given as:

y − b = − 1/m(x − a) The equation of the normal is y − k = m(x − h) Where (a, b) is a point on the circumference of the circle and (h, k) is the centre of the circle.

Example 4.9

Find the equations of the line tangent and normal to the circle x²+ y²= 169 at point (5, − 12)

Solution

First find the centre of the circle.

To do this, we will write the equation in the standard form:

(x − 0)²+ (y − 0)²= 13²This is a circle with a centre (0, 0) and a radius of 13 units.

Next, we will check if the given point satisfies the circle equation.

5²+ ( − 12)²= 25 + 144 = 169 This shows that (5, − 12) is on the circumference of the circle.

Next, find the gradient of the normal. This is the gradient of the line connecting the points (0, 0) and (5, -12) m = −12 − 0_ 5 − 0 = − 1 2_ 5 Equation of the normal is: y − 0 = −12/5 (x − 0) Multiply through by 5:

5y = − 12x 5y + 12x = 0 is the equation of the normal.

The gradient of the tangent is = − 1/m = − 1__ 12/5 = 1 × 5/12 = 5/12 The equation of the tangent is: y − (− 12) = 5/12(x − 5) (y + 12) = 5/12(x − 5) Multiply through by 12:

12y + 144 = 5x − 25 12y − 5x + 144 + 25 = 0 12y − 5x + 169 = 0 is the equation of the tangent.

Example 4.10

The point (5, 3) satisfies the equation x²+ y²− 6x − 4y + 8 = 0 .

a. Find the radius of the circle.

b. Fine the equation of the normal to the circle at point (5, 3)

c. Find the equation of the tangent to the circle at (5, 3)

Solution

a. To find the radius, write the equation in standard form x²− 6x + 9 + y²− 4y + 4 = − 8 + 9 + 4 (x − 3 )²+ (y − 2 )²= 5 (x − 3 )²+ (y − 2 )²= (√5)²This means the radius is √5 and the centre is (3, 2).

Gradient of the normal.

This is the gradient of the line connecting the points (3, 2) and (5, 3)

b. The gradient of the normal = 3 − 2/5 − 3 = 1/2 The equation of the normal is: y − 2 = 1/2(x − 3) 2y − 4 = x − 3 2y − x − 4 + 3 = 0 2y − x − 1 = 0 Gradient of the tangent = − 1 ÷ 1/2 = − 2

c. The equation of the tangent is:

y − 3 = − 2(x − 5) y − 3 = − 2x + 10 y + 2x − 13 = 0

Example 4.11

Find the equation of the normal to the circle y²+ x²+ 8x + 7 = 0 that passes through the point (1, 3)

Solution

First, verify if the point lies on the circumference of the circle:

(3)²+ (1)²+ 8(1) + 7 ≠ 0 Therefore, the point (1, 3) is not on the circle’s circumference.

Next, find the coordinates of the centre of the circle.

The centre is given as = (coefficient of x___________

–2 , coefficient of y___________ − 2 ) = ( 8___ − 2, 0___ − 2) = ( − 4, 0) The gradient of the normal = − 3______ − 4 − 1 = 3/5 Equation of the normal: y − 3 = 3/5(x − 1) 5y − 15 = 3x − 3 5y − 3x − 15 + 3 = 0 5y − 3x − 12 = 0 The diagram below shows the construction of the circle and the normal

Figure 4.23: Construction of a circle and normal Length of a tangent (L) Given the equation of a circle, it is possible to find the length of a tangent to the circle from a point (x, y) outside the circle.

Figure 4.24 illustrates this:

Figure 4.24: Length of a tangent In Figure 4.24, V(a, b) is the centre of the circle. T is a point of tangency and P(x, y) is any point outside the circle. The length of the tangent is |PT| = L .

From Pythagoras’ theorem, we have L²+ r²= d², since we are interested in finding L, make it the subject.

L²= d²− r²√L²= √d²− r²L = √d²− r²Example 4.12 Find the length of the tangent from the point (0, 4) to the circle x²+ y²− 2x + 12y − 3 = 0

Solution

First find the centre and radius of the circle:

x²− 6x + 9 + y²+ 4y + 4 − 3 = 9 + 4 (x − 3 )²+ (y + 2 )²= 16 (x − 3 )²+ (y + 2 )²= 4² The centre of the circle is (3, -2) and the radius is 4

Figure 4.25: Length of the tangent to the circle x²+ y²− 2x + 12y − 3 = 0 Distance between the (0, 4) and the centre (3, -2), d = √_____________ (3 − 0)²+ (− 2 − 4)²d = √___________ (3 )²+ ( − 6 )²d = √9 + 36 d = √45 Length of the tangent = √d²− r²Length of the tangent = √__________ (√45)²− (4)²= √45 − 16 = √29

Example 4.13

Figure 4.26 shows a graph of a circle. Line OBA is a tangent to the circle at B.

Figure 4.26: Graph of a circle with Line OBA as tangent

a. Find the equation of the circle.

b. Find |AB|, leave your answer in surds

Solution

a. The centre of the circle is (-5, -2) and the radius is 3.

Figure 4.27: Graph of a circle with Line OBA as tangent The equation of the circle is: (x − ( − 5 )²+ (y − ( − 2) )²= 3²(x + 5 )²+ (y + 2 )²= 9 x²+ 10x + 25 + y²+ 4y + 4 − 9 = 0 x²+ y²+ 10x + 4y − 5 = 0

b. Coordinates of point A is (1, -2) Distance between point A and the centre (-5, -2), d = √________________ (− 5 − 1)²+ (− 2 − ( − 2))²d = √___________ ( − 6 )²+ (0 )²d = √36 = 6 d = 6 Length of the tangent = √6²− 3²= √36 − 6 = √30

Locus A locus (plural = loci) is a set of points which satisfies a given condition or criterion. The result of a locus will either be a complete circle, a part of a circle (arc), a line or a single point. Here, we will learn about some conditions or criteria a set of points satisfies.

Condition 1:

The Locus of points are equidistant from one point.

The result of this locus is a circle which will be r units away from the given point.

The given point is the centre of the circle. The criterion or the condition is that:

“it is equidistant from the centre”

Figure 4.28 shows the locus of all points r units from point (h, k)

Figure 4.28: Locus of all points r units from point (h, k)

Example 4.14

Find the equation of all points 4 units away from ( − 1, 2)

Solution

This is the same as finding the equation of a circle with centre (− 1, 2) and radius = 4.

Using the general equation of the circle, we have:

(x − (− 1) )²+ (y − 2 )²= 4²(x + 1 )²+ y²− 4y + 4 = 16 x²+ 2x + 1 + y²− 4y + 4 = 16 x²+ y²+ 2x − 4y + 1 + 4 − 16 = 0 x²+ y²+ 2x − 4y − 11 = 0 is the locus of all points 4 units away from the point ( − 1, 2) Condition 2:

The locus of points equidistant from two fixed points.

The result of this locus is a line which will perpendicularly bisect the line joining the two points.

Figure 4.29 below shows the locus of points, L, equidistant from points P and Q.

Line L bisects PQ perpendicularly at m, the midpoint of line PQ.

Figure 4.29: Locus of points, L, equidistant from points P and Q

Example 4.15

Calculate the equation of all points equidistant from A(-5, 0) and B(3, -6)

Solution

Midpoint of line AB = (− 5 + 3/2 , 0 + − 6/2 ) = (−2/2, −6/2) = ( − 1, − 3) Gradient of line AB = − 6 − 0/3 — 5 = −6/8 = −3/4 Gradient of the locus = − 1 ÷ (−3/4) = − 1 × −4/3 = 4/3 Equation of the locos is:

y − − 3 = 4/3(x − ( − 1)) y + 3 = 4/3(x + 1) Multiply through by 3:

3 × (y + 3) = 3 × 3/4 (x + 1) 3y + 9 = 4(x + 1) 3y + 9 = 4x + 4 3y − 4x + 9 − 4 = 0 3y − 4x + 5 = 0 Therefore, 3y − 4x + 5 = 0 is the locus of all points equidistant from (-5, 0) and (3, -6) Condition 3:

The locus of points equidistant from two intersecting lines.

The result of this locus is a pair of angle bisectors. For any point (x, y) on the angle bisector, the perpendicular distance is equidistant from the two lines.

In Figure 4.30, l and m are the loci equidistant from line AB and line CD.

Line L bisects angle BOD and M bisects angle AOD.

Also, for any point P(x, y) on the locus, L, the perpendicular distance between point P and line AB is equal to the perpendicular distance between point P and line CD.

Figure 4.30: Locus of points equidistant from two intersecting lines

Example 4.16

Find the equation of the locus of points which moves so that its distance from the lines y = 2x − 3 and x + 2y = 5 is the same.

Let P(x, y) be any point on the loci. This means that the perpendicular distance between P(x, y) and the line y = 2x − 3 is equal to the perpendicular distance between point P(x,y) and the line x + 2y = 5 Recall that the perpendicular distance between a point (m,n) and a line ax + by + c = 0 is |ᵃ(m) + b(n) + c___________ √a²+ b²| The equation of the loci is given by:

| ʸ− 2x + 3_ √1²+ (− 2)²| = |²ʸ+ x − 5_ √2²+ 1²| |ʸ− 2x + 3_ √5 | = |²ʸ+ x − 5_ √5 | multiply both sides by √5, which effectively removes the denominator:

|ʸ− 2x + 3| = |2ʸ+ x − 5| Remove the absolute value sign by multiplying any side of the equation by ± In this example, we chose to remove the absolute value sign by multiplying the RHS of the equation by ± y − 2x + 3 = ±(2y + x − 5) So, either y − 2x + 3 = 2y + x − 5 0 = 2y − y + x + 2x − 5 − 3 y + 3x − 8 = 0 or y − 2x + 3 = − 2y − x + 5 y + 2y − 2x + x + 3 − 5 = 0 3y − x − 2 = 0 Therefore, the equation of the locus is either y + 3x − 8 = 0 or 3y − x − 2 = 0 Condition 4:

The locus equidistant from a line.

The result of this locus is a pair of lines equidistant from the given line.

For instance, the locus of points d units away from line AB is either line M or line L as shown in Figure 4.31.

Figure 4.31: Locus equidistant from a line So, if the equation of AB is ax + by + c = 0, then d = |ᵃˣ+ by + c_________ √a²+ b²|

Example 4.17

Find the locus of the point R(x, y) such that it is always 3 units from the line y + 1 = 0

Solution

3 = |⁰ˣ+ y + 1_ √0²+ 1²| 3 = |ʸ+ 1_ √1 | 3 = ±|ʸ+ 1| 3 = y + 1 or 3 = − y − 1 y − 2 = 0 or y + 4 = 0 Therefore, the locus R(x, y) of points which is 3 units away from y + 1 = 0 is either y − 2 = 0 or y + 4 = 0

Example 4.18

If P(1, 4) and Q(5, -1) are two fixed points and A(x,y) moves such that |AP|:|AQ| = 1:2. Find the equation of the locus.

Solution

From the condition given, |AP| : |AQ| = 1 : 2 |AP|_ |^(AQ)| = 1/2 2|AP| = 1 × |^(AQ)| 2 √____________ (y − 4)²+ (x − 1 )²= √_______________ (y − ( − 1))²+ (x − 5 )²2 √____________ (y − 4)²+ (x − 1 )²= √____________ (y + 1)²+ (x − 5 )²Square both sides:

4[y²− 8y + 16 + x²− 2x + 1] = y²+ 2y + 1 + x²− 10x + 25 4 y²− 32y + 64 + 4 x²− 8x + 4 = y²+ 2y + 1 + x²− 10x + 25 4 y²− y²− 32y − 2y + 4 x²− x²− 8x + 10x + 68 − 26 = 0 3 x²+ 3 y²+ 2x − 34y + 42 = 0 is the equation of the locus.

Example 4.19

A point A(x, y) moves such that AC and AD are perpendicular.

If C(-3, 0) and D(3, 5), find the equation of the locus A.

Solution

Since AC and AD are perpendicular, (gradient of AC)× (gradient of AD)= − 1 ( y − 0_ x − (− 3)) × (y − 5_ x − 3) = − 1 ( y_ x + 3) × (y − 5_ x − 3) = − 1 y²− 5y_ x²− 9 = −1_ 1 Cross multiply:

y²− 5y = − 1( x²− 9) y²− 5y = −x²+ 9 y²+ x²− 5y − 9 = 0 is the locus A

1. The diagram below shows the graph of three circles, P, Q and R.

2. Use it to answer the questions.

a. State the coordinates of the centre of P, Q and R.

b. Find the radii of circles P, Q and R

c. Write the equation of circle R in general form

d. Write the equation of circle P in the standard form

e. Write the equation of circle Q in the standard form

f. Find the perimeter of the triangle formed by the centres of the three circles.

g. Give your answer correct to three significant figures.

3. Write the general form of the circle equation with the following properties.

a. Centre (1, -1) and radius 7

b. Centre (-2, 1) and radius 5

c. Centre (0, -3) and radius 2

d. Centre (-3, -4) and radius 8

4. The diagram shows the graph of circles A, B and C.

Use them to answer the following questions.

a. Write the equation of each circle in the general form

b. Find the perimeter of the triangle formed by the centres of the circle.

Leave your answer in surd form.

5. A TV remote has a range of 8m in all directions. If the TV is 5m East and 3m South, write an equation to represent the area within which the TV receives signals from the remote. Can the TV receive signals from the remote if the remote is located 2m West? Justify your response.

6. An Uber driver operates within 30km of a post office at all angles. His house is 5km West and 12km south of a post office.

7. a. Write an equation to represent the driver’s travel boundary from the post office using the driver’s house as the origin. Leave your equation in standard form.

b. Find the furthest distance between the driver’s house and his travel boundary.

c. Find the shortest distance between the driver’s house and the driver’s travel boundary.

8. Write the general equation of a circle with:

a. Centre (4, − 1) and radius √11.

b. Centre ( − 2, − 3) and radius 2 1/2

c. Centre (0, − 3 ) and radius 1/3

d. Centre (1/2, − 1) and radius 1 1/2

9. Write each of the following circles in the standard form. Also, state the centre and radius of each circle.

a. x²+ y²− 2x + 2y − 47 = 0

b. x²+ y²+ 6y = 0

c. x²+ y²− 12x − 64 = 0

d. x²+ y²+ 4x − 8y + 4 = 0

e. 2x²+ 2 y²− 2x − 2y − 7 = 0

f. 3 x²+ 3 y²+ 12x − 2y + 4 = 0

10. The endpoints of the diameter of a circle are (0, 5) and (0, -7).

11. Find the

a. radius and centre of the circle

b. general equation of the circle.

12. The endpoints of the diameter of a circle are (6, 0) and (-2, 0).

13. Find the

a. radius and centre of the circle

b. general equation of the circle.

14. The endpoints of the diameter of a circle are (-4, 3) and (4, -3).

15. Find the

a. radius and centre of the circle

b. general equation of the circle.

16. In each of the following questions, coordinates of three points are provided.

Write the general equation of the circle that satisfies all three points. State the centre and radius of the resulting circle.

a. ( − 5, 0), ( − 1, − 4) and ( − 1, 4)

b. (6, − 1), (3, − 4) and (0, − 1)

17. Find the equations of the line tangent and normal to the circle.

a. x²+ y²= 100 at point ( − 8, − 6)

b. x²+ y²= 49 at point (7, 0)

c. x²+ y²− 2x − 24 = 0 at (1, -5)

18. Find the length of the tangent from the point:

a. (-8, 6) to the circle x²+ y²− 25 = 0 , leave your answer in surd form

b. (3, 6) to the circle x²+ y²− 4y − 5 = 0

c. (7, − 3) to the circle x²+ y²+ 10x − 4y + 13 = 0 , leave your answer in surd form

d. (5, −1/2) to the circle 4x²+ 4 y²+ 12x + 4y − 15 = 0

19. The diagram below shows a graph of a circle. Line BA is a tangent to the circle at B.

a. Find the equation of the circle.

b. Find |AB|, leave your answer in surds

20. Calculate the equation of all points equidistant from A(-4, 0) and B(2, 8))

21. Find the equation of the locus of points which moves so that its distance from the lines 3y − x = 0 and x − 3y = 0 is the same.

22. Find the locus of point R(x, y) that is always 2 units from the line x + 5 = 0

23. If P(-2, 1) and Q(4, -1) are two fixed points and A(x, y) moves such that |AP|:|AQ| = 2:1. Find the equation of the locus.

24. A point A(x, y) moves so that AC and AD are perpendicular. C(-2, 0) and D(6, 4).

25. Find the equation of the locus A.

Learning about vectors and their projections is important because vectors allow us to understand and describe quantities that have both magnitude and direction such as forces, velocities and electric fields. In this section you will learn about transposing vectors, dividing a vector internally based on a ratio and finding the dot (scalar) product of vectors. You will also learn about finding angles between two vectors, the application of the sine and cosine laws and the projection of one vector onto another.

KEY IDEAS

• In transposing a vector, change the orientation from a column to a row and vice versa.

• The angle between two vectors, a and b, is given by θ in cos θ = a . b____ |a||b| o where a . b = a₁ b₁ + a₂ b₂ (in two dimensions) = a₁ b₁ + a₂ b₂ + a₃ b₃ (in three dimensions).

• The vector ⟶ AB is given by b – a where a and b are the position vectors of A and B.

In year one we learnt about straight lines and their properties (section 5), vectors (section 6) and trigonometry (section 7). In this section we will make use of the stated concepts learned in year one to help us understand and solve problems in relation to vectors. Let us work through this activity to refresh our minds on lines and vectors.

Activity 5.1: Treasure Hunt on the Coordinate Plane (in small groups, pairs or individually)

1. Imagine you are on a treasure hunt! Start by marking the point (0, 0) on your graph paper as your starting point. Each vector will guide you closer to the treasure!

2. Follow the Directions:

a. Move from (0, 0) to the first position with ⟶ OA = (3 4)

b. Add the vector ⟶ OB = (− 2 5 ) to get to the next position

c. Add the vector ⟶ OC = ( 4 − 3)

d. Subtract the vector ⟶ OD = (− 3 − 4) to go to the next position

e. Add the vector ⟶ OE = (1

2) for your final move

f. After each move, mark your new spot and draw a line from your last point to your current one.

3. Congratulations! You’ve found the treasure.

4. Write out the coordinates of the treasure position. Check that your finishing position is the same as your classmates.

5. Look for Patterns

a. Check out your line segments:

i. Notice the gradient for each line segment and see if any gradients (slopes) are the same or opposite.

ii. Are any segments parallel or perpendicular?

iii. Observe the distance between each point you marked.

b. Do the segments form any special shapes, like triangles or parallelograms?

6. How did adding and subtracting vectors help you reach the treasure?

7. What straight-line properties did you notice along the way?

Now that we’ve gone through this activity, let us talk about transposing vectors!

Transposition simply means to change something from one position to another.

In transposing a vector, you change the orientation from a column vector to a row vector or change a row vector to a column vector.

For example, the vector, W = ⎛ ⎜⎝ w₁ w₂ w₃ w₄ .

.

.

wₙ ⎞ ⎟⎠ when transposed becomes Wᵀ= (w₁ w₂ w₃ w₄… wₙ).

Example 5.1

Transpose the vector M = ( 2 5 − 7).

Solution

Step 1: Identify whether the given vector is a column or row vector.

In this case it is a column vector

Step 2: Identify the vector orientation it should change to.

In this case, a row vector

Step 3: Rearrange the elements of the vector to suit the new orientation.

Mᵀ= (2 5 − 7)

Example 5.2

Transpose the vector H = (− 5 6 8 3).

Solution

Step 1: Identify whether the given vector is a column or row vector.

Row vector

Step 2: Identify the vector orientation it should change to.

Column vector

Step 3: Rearrange the elements of the vector to suit the new orientation.

Hᵀ= ⎛ ⎜ ⎝ − 5 6 8 3 ⎞ ⎟ ⎠

In year one, we learnt how to divide straight lines internally based on a ratio and how to extend a line segment based on a ratio. We will apply this knowledge to divide a vector in a given ratio.

Let us use the concept of unit vectors to assist in dividing vectors through these examples.

Example 5.3

R and T are points on the position vectors 8i + 3j and − 3i + 5j , respectively on a vector ⟶ RT, S is a point on ⟶ RTsuch that |RS| : |ST| = 3:1. Find the position vector of the point S.

Solution

Figure 5.1: Dividing (⟶

RT) by ratio 3:1 ⟶ OS = 1(⟶ OR) + 3( ⟶ OT)____________ 3 + 1 = 1(8i + 3j) + 3( − 3i + 5j)__________________ 4 = 8i + 3j − 9i + 15j/4 = −i + 18j_ 4 = − 0.25i + 4.5j

Example 5.4

A and E are points on the position vectors − 6i + 5j and 2i − 9j , respectively on a vector ⟶ AE, B is a point on ⟶ AE such that |AB| : |BE| = 7:3. Find the position vector of the point B.

Figure 5.2: Division of ⟶ AE in the ratio 7:3

Solution

⟶ OB = 3(⟶ OA) + 7( ⟶ OE)_____________ 3 + 7 = 3(− 6i + 5j) + 7(2i − 9j)__________________ 10 = −18i + 15j + 14i− 63j/10 = −4i− 48j_ 10 = − 0.4i− 4.8j

As the name implies anything about products is to do with multiplication. In this case, the dot product of vectors focuses on multiplying the corresponding coordinates of given vectors and summing the results.

For example, given the vectors m = ( a₁ b₁) and n = ( a₂ b₂) the dot product will be m . n = (a₁ × a₂)+ (b₁ × b₂) which will result in = a₁ a₂+ b₁ b₂

Note that m .n is read as “m dot n ” Let us go through this activity to explore what the dot product of a vector and its transposed vector will be.

Activity 5.2: Investigating Dot Products (working individually or in pairs)

1. Choose a vector (e . g. m = ( u₁ v₁) )

2. Transpose the vector (e . g. m T )

3. Find the dot product of the original vector and its transposed vector

4. Discuss your results with a classmate.

Hopefully you found, in this case, that m. mᵀ= (u₁ × u₁) + (v₁ × v₁) = (u₁)²+ (v₁)²= m. m Let’s go ahead to explore how an angle between two vectors relates to the dot product of the vectors.

Activity 5.3: Investigating Angles Between Vectors (working in small groups)

1. Create two vectors of your choice.

2. Represent these vectors on a graph (manually or electronically).

3. Join the vectors from a third point, the origin.

4. Measure the angle formed between the two vectors at the origin.

5. Find the cosine of the angle formed and document it.

6. Now find the magnitude of each of the two vectors.

7. Find the product of the two magnitudes and the documented result in

step 5

8. At this point, compute the dot product of the two vectors.

9. Compare the results in steps 7 and 8.

10. What conclusion can be drawn from the results?

Hopefully you concluded that the dot product is the same as the product of the two magnitudes and cosine of the angle between the vectors.

Generalisations

1. m .n = |m||n|cosθ

2. cosθ = . n____ |m||n|

3. m .n = n .m (Dot product is commutative)

Activity 5.4: Algebraic Proof of the Dot Product (work in groups) Carry out research on the algebraic proof for a .b = |a||b|cosθ.

Example 5.5

Given the vectors t = ( 7 − 3) and k = (5 8), find the dot product of t and k.

Solution

Step 1: Identify corresponding coordinates 7 and 5, − 3 and 8

Step 2: Multiply the identified pairs and find their sum:

t . k = (7 × 5) + ( − 3 × 8) = 35 + ( − 24)

Step 3: Simplify the results:

t . k = 35 − 24 = 11 Therefore, t . k = 11 .

Example 5.6

Given the vectors p = (1

4) and q = (− 2 3 ). Find the:

a. dot product

b. angle formed between p and q

Solution

a. p . q = (1 × (− 2)) + (4 × 3) = 10

b. cosθ = . q____ |p||q|,

Step 1: Make θ the subject.

θ = cos⁻¹( p . q_ |p||q|)

Step 2: Find the magnitude of p and q .

|p| = √__________ (1)²+ (4)²= √17 |q| = √__________ ( − 2)²+ (3)²= √13

Step 3: Substitute the values into θ = cos⁻¹( p . q____ |p||q|).

θ = cos⁻¹( ¹⁰_ ( √17)( √13)) θ = 47.7263⁰You can confirm the size of the angle by plotting the vectors and measuring the angle using GeoGebra Software.

Properties of the scalar (dot) product

1. The dot product of any two vectors is a scalar and not a vector, hence it can be known as the scalar product, rather than dot product.

2. The dot product of a vector with itself is the square of its magnitude.

a . a = |a|²3. Parallelism Property In year one we learnt about parallel vectors. If you remember, parallel vectors are scalar multiples of each other and do not intersect at any point.

This helps to establish that no angle can be formed between two parallel vectors. Hence, u .v = |u||v|cos(0) but cos(0) = 1 leading to;

u .v = |u||v| for parallel vectors.

So, the dot product becomes same as the product of the magnitudes for parallel vectors.

Curious about the validity of this property? Choose parallel vectors and explore using GeoGebra software or manually.

4. Perpendicular Property

You know that cos 90° = 0 and perpendicular vectors have an angle of 90° formed between them.

Therefore, it can be said that if vectors u and v are perpendicular then u .v = 0. Also, if the scalar product of two non-zero vectors is zero, they must be perpendicular.

5. Multiplication by a constant ma ∙ nb = mn|a ||b | where m and n are constants and a and b are vectors

6. The scalar(dot) product is distributive over addition.

Hence a ∙ (b + c) = a ∙ b + b ∙ c where a , b and c are vectors.

The sine and cosine rules are very useful. They provide ways of finding angles and lengths when the triangle is not necessarily right-angled.

Let’s go through this activity to investigate these laws.

Activity 5.5: Investigating the Cosine and Sine Rules (work in small groups)

1. Draw three different triangles (1 acute, 1 obtuse, 1 right angled triangle)

2. Label the vertices of each triangle as Q, R, and T.

3. Label the side opposite each angle as q, r, and t respectively (ie, q is opposite Q, r is opposite R, and t is opposite T).

4. Use a ruler to measure each side q, r and t in centimetres (cm).

5. Use a protractor to measure each angle Q, R, and T in degrees (°).

6. Record your measurements in a table like this:

Triangle type q(cm) r(cm) t(cm) Q(°) R(°) T(°) Acute Obtuse Right-angled

7. For each triangle, check if t²= q²+ r²− 2qr cos(T) holds.

8. For each triangle, calculate and record the following ratios:

q_ sinQ, r___ sinR, t___ sinT

9. What do you notice about the relationships?

10. Does the equation with cos hold true for all the triangles?

11. Does the equation with sin hold true for all triangles?

12. Discuss with a classmate or write a short paragraph explaining what this relationship reveals about any triangle. Consider why this must be useful in maths.

13. These are the Cosine and Sine Rules.

Generalisations:

1. The Cosine Rule states that, in any triangle, the length of one side can be calculated if we know the lengths of the other two sides and the angle between them:

t²= q²+ r²− 2qr. cos(T)

2. The Sine Rule helps us find missing sides or angles in any triangle if we have some known values:

q_ sinQ = r____ sinR = _ t___ sinT .

Example 5.7

Two boats named Triumph and Mayflower leave a harbour at the same time.

Triumph heads east and travels 5km while Mayflower heads north-east and travels 8km. Find the distance between the two boats after the journey.

Solution

Step 1: Make a rough sketch of the positions and movement of the boats.

Figure 5.3: Representation of boat movements

Step 2: Choose a variable to represent the distance between the two boats.

Let d be the distance between the two boats.

Step 3: Identify the angle formed between the boats α = 45⁰Step 4: Apply the cosine rule:

d²= 5²+ 8²− 2(5)(8). cos(45) d²= 89 − 40 √2 = √32.43 = 5.69km

Example 5.8

Two observation towers are located 500 metres apart along a coastline at points T₁and T₂. From T₁ a ship is sighted at a bearing such that the angle between the line T₁ T₂ and T₁ S is 40⁰. From T₂ the ship is sighted at a bearing such that the angle between T₂ T₁ and T₂ S is 75⁰. Find the distance between the tower T₁ and the ship using the sine law.

Solution

Step 1: Make a rough sketch of the positions and points of the Tower.

Figure 5.4: Representation of Towers and ships

Step 2: Since we have two angles, determine the third angle in the triangle.

< T₁ S T₂ = 180⁰− 40⁰− 75⁰= 65⁰Step 3: Apply the sine law ST₁_ sin 75⁰= T₁T₂_____ sin65°

Step 4: Substitute the value for T₁ T₂ and simplify the equation ST₁_ sin 75⁰= 500_____ sin65⁰ST₁ = sin 75⁰(500)__________ sin65⁰ST₁ = 532.89 metres Therefore, the distance between tower T₁ and the ship is 532.89 metres.

Area of a triangle using the Sine ratio You already know that area of triangle is 1/2 base × perpendicular height? We can use this information to find the area of triangles using the sine rule.

Let us work through this activity to find out how.

Activity 5.6

1. Draw an isosceles or scalene triangle.

2. Label the points of the triangle.

3. Indicate the perpendicular height of the triangle.

Figure 5.5: Exploring area of a triangle

4. Express the angle β in terms of the height using the sine trigonometry identity.

For instance, sinβ = ||AD|____ |AC| .

5. Now make the perpendicular height the subject (in this case, |AD| = |AC |sinβ).

6. From our triangle, the base is |BC|, now substitute the base and height into the known formula for area of a triangle.

7. So, we have: 1/2 × |^(BC)| × |AC|sinβ

8. This is area of a triangle in terms of sine and it does not matter if we do not have the perpendicular height making it very useful.

Generalisation The area of a triangle equals a half of the product of the magnitudes of the lengths of two sides (or two vectors which bound the triangle) and the sine of the angle between the two known sides (or vectors).

Let’s go through these examples to consolidate knowledge.

Example 5.9

If Q(2, -3), R(6, 2) and T(-4, 1), are the vertices of ∆ QRT , calculate:

a. <QRT to two significant figures.

b. the area of triangle QRT to two significant figures.

Solution

Figure 5.6: Graphical Representation of example 5.9

Step 1: Represent <QRT by θ

Step 2: Find the dot product of ⟶ RQ and ⟶ RT and make the angle the subject.

⟶ RQ =( 2 − 6 − 3 − 2), ⟶ RT = (− 4 − 6 1 − 2 ) ⟶ RQ = ( − 4 − 5 ), ⟶ RT = (− 10 − 1 ) ⟶ RQ . ⟶ RT = |⟶ RQ| |⟶ RT| ^(cosθ)40 + 5 = √__________ ( − 4)²+ ( − 5)²× √___________ ( − 10)²+ ( − 1)²cosθ 45 = 64.3506 cosθ θ = cos⁻¹( 45_ 64.3506) θ = 45.63⁰θ ≈ 46⁰to two significant figures.

Step 3: Use the sine rule to find the area Area = 1/2 × |^(RQ)| × |RT|sinθ = 1/2 × √41 × √101 × sin 46⁰= 23.1450 The area of triangle QRT is 23 square units to two significant figures.

Do you know how a projector works? When we project a presentation or movie onto another screen, we allow for what is on the original screen (e.g., computer) to be seen on the projected screen (e.g. whiteboard). An image of whatever is on the original screen is projected onto a new screen.

According to the Cambridge dictionary, a projection is a calculation or guess about the future based on information you have. In studying vectors, we can also project one vector onto another vector allowing us to determine the shortest distance between that point and the line.

When you project one vector onto another vector, you can either compute the scalar projection or the vector component.

Let us go through this activity to see how the projection works.

Activity 5.7 – Exploring Vector Projections

1. Draw a vector A from O going up and to the right, at a 45° angle, about 6 cm long.

2. Draw a second vector B from O going horizontally to the right (at a 0° angle), about 8 cm long.

3. Label both vectors and mark their directions with arrows.

4. Use your protractor to measure the angle between vector OA and vector OB.

5. Label this angle as θ.

6. Imagine shining a light straight down on vector A. The shadow of A will fall along B.

7. Imagine OA as a shadow, how much of OA falls along OB

8. The goal is to draw how long this shadow would be if A was stretched along the direction of B.

9. Place your ruler along vector OB (the horizontal line).

10. From the tip of A, draw a straight line down to B, making sure this line is at a right angle (90°) to B.

11. Where this line hits B is the “shadow” tip of A on B.

12. Measure the distance from point O to the point where the shadow hits B.

13. This distance is the projection of A onto B.

14. The distance shows you how much of A falls along the same direction as B.

Figure 5.7: Projection of ⟶ OA on ⟶ OB

15. Your diagram should look like Figure 5.7: Projection of ⟶ OA on ⟶ OB.

Generalisation

1. The scalar projection of a vector a on vector b is, Proj_(b)a = a . b____ |b| or |a|cosθ

2. The vector component of a projection of a vector a on a vector b is a . b____ |b| ( ˆb ) or (ₐ . ˆb ) ˆb Let’s go through these examples to see how to calculate the projection of a vector on another vector.

Example 5.10

A farmer in Northern Ghana is setting up irrigation pipes. Due to wind blowing in the direction of W = (4 3), she needs to align the water flow direction vector I = (6

8) so that it’s not significantly affected by the wind. Calculate the projection of the irrigation flow vector I onto the wind vector W and interpret.

Solution

Step 1: Identify the vectors.

W = (4 3), I = (6 8)

Step 2: Compute the dot product of the vectors.

I.W = (4 × 6) + (3 × 8) = 24 + 24 = 48

Step 3: Find the magnitude of W |w| = √(4)²+ (3)²|w| = 5

Step 4: Calculate the scalar projection of I on W Proj_(w)I = I . W____ |w| = 48/5 = 93/5 = 9 . 6

Step 5: Write out the interpretation The irrigation flow vector has a 9.6-unit component in the direction of the wind vector. This means that the wind will push the irrigation flow slightly off its intended path in the direction of the wind by 9.6 units. Hence the farmer should adjust the angle of the irrigation pipe slightly to counteract this drift caused by the wind.

Example 5.11

A construction worker is using a ramp to transport building materials onto a platform. The ramp is inclined at an angle of 30° to the ground and the force required to move a load along the ramp is measured to be 100N. Calculate the effective force acting in the direction of the horizontal that is needed to move the load.

Solution

Step 1: Identify the angle of inclination.

θ = 30°

Step 2: Identify the magnitude of the Force needed.

|F| = 100N

Step 3: Calculate the scalar projection |F|cosθ = 100cos 30° = 86.60 N The force acting in the direction of the horizontal that is needed to move the load upwards is 86.60 N.

1. A boat is positioned at point A (0, 0) on a river, and another point B(6, 0 ) is located 6 km downstream. A third point P on the river, representing a midway check, is supposed to divide AB internally in the ratio 2:3.

a. Determine the coordinates of point P .

b. If the boat travels from A to P at a speed of 2 km/h, calculate the time it takes to reach P .

2. Relative to a fixed origin O, the respective position vectors of three points A, B and C are (3 2), (− 4 4 ) and (4 0).

a. Determine, in component form, the vectors ⟶ AB and ⟶ AC

b. Hence find, to the nearest degree, the angle BAC.

c. Calculate the area of the triangle BAC.

3. Relative to the position vector O, the position vectors are given by:

4. ⟶ OA = ( 1 6 11), ⟶ OB = ( 4 3 5), ⟶ OC = ( 6 1

1) and ⟶ OD = ( 2 8 9).

a. Show clearly that ⟶ AD is perpendicular to ⟶ BD

b. State the ratio AB: BC

c. Determine the area of triangle ABD

5. Find the angle between the vectors. Give your answers to 1 decimal place.

a. 2i + 3j and 4i + j

b. —i — j and —i —2j

c. (2

3) and (6 4)

6. Relative to an origin O, the position vectors of the points A and B are given by

7. ⟶ OA = 2i − 8j + 4k and ⟶ OB = 7i + 2j − k.

a. Find the value of ⟶ OA . ⟶ OB and hence state whether angle AOB is acute, obtuse or a right angle.

b. If the point X is such that ⟶ AX= 2/5 ⟶ AB, find the unit vector in the direction of OX.

8. Miriam has her own small helicopter. One afternoon, she flies for 1 hour with a velocity of 120i + 160j km/h where i and j are unit vectors in the directions east and north. Then she flies due north for 1 hour at the same speed. Finally, she returns to her starting point; flying in a straight line at the same speed. Find, to the nearest degree, the direction in which she travels on the final leg of her journey and, to the nearest minute, how long it takes her.

9. An aircraft travels from City A(0, 0) to City B(250, 0), which are 250 km apart. A fuel check point, F is located at a point dividing AB internally in the ratio 1:4, closer to City A .

a. Find the coordinates of the fuel check point F .

b. If the airplane flies from City A at a speed of 500 km/h, how long will it take to reach F ?

10. Vectors b and e are such that |b|= 5cm, |e|= 12cm and |b + e|= √229. Find

a. The angle between b and e.

b. The scalar (dot) product of b and e.

11. Transpose the vector g = (15 6 7).