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Grade 10 Mathematics: Learners’ Book

INNODEMS

Section 2.8 Vectors I

In mathematics, vectors are a fundamental concept that goes beyond numbers. Suppose I ask, "How far is your home from school?" One possible response is " \(2\) kilometers." However, I can’t get to your home with just this information. I would also need to know the direction, whether it is east, west, northeast, or south. This combination of both distance and direction is what a vector represents.
Pilots use vectors to calculate the distance and direction they need to take inorder to travel from one location to another. In this section, we will explore how to use vectors and apply different operations on them.

Subsection 2.8.1 Vector and Scalar Quantities

Activity 2.8.1.

Work in groups
(a)
Identify and list at least five examples of quantities that have both magnitude and direction.
(b)
Identify and list at least five examples of quantities that have only magnitude.
(c)
Discuss how these quantities are used in real life scenarios.
(d)
Discuss and share your findings with the rest of the class.
\(\textbf{ Key Takeaway } \)
A vector is a quantity that has both magnitude and direction, whereas a scalar is a quantity which has only magnitude.
Examples of vector quantities include force, velocity, and displacement, while scalar quantities include mass,temperature, and speed etc.
Table 2.8.1. Difference Between Vector and Scalar Quantities
Feature Vector Quantity Scalar Quantity
Definition Has both magnitude and direction Has only magnitude
Examples Force,Acceleration,Displacement Distance,Temperature,Mass

Subsection 2.8.2 Representation of Vectors and Vector Notation

Activity 2.8.2.

Work in groups
(a)
Go into the school compound field, choose a starting point and mark it as point \(\text{A}\text{.}\)
(b)
Standing at point \(\text{A}\text{,}\) locate the north, south, west, and east directions.
(c)
From point \(\text{A}\text{,}\) walk \(20\) steps to the north direction and mark it as point \(\text{B}\text{.}\)
(d)
From point \(\text{B}\text{,}\) walk \(15\) steps to the east direction and mark it as point \(\text{C}\text{.}\)
(e)
On a piece of graph paper, draw a sketch that shows your path from point \(\text{A}\) to \(\text{B}\) and then from \(\text{B}\) to \(\text{C}\text{.}\)
(f)
Think of a way on how you can represent the movement from point \(\text{A}\) to \(\text{B}\) using notation, also from point \(\text{B}\) to \(\text{C}\text{.}\)
(g)
Suppose you now move in a reverse way from point \(\text{C}\) to point \(\text{A}\) following the same path. How would you represent that movement using notation?
(h)
Discuss your ideas and representations with the rest of the learners in class.
\(\textbf{ Key Takeaway }\)
We can represent a vector by a line drawn between two points as shown in Figure 2.8.2 below:
Figure 2.8.2.
Vector notation is a way of representing quanties that have both magnitude and direction. Vector \(\textbf{PQ}\) can be denoted as \(\underset{\sim}{PQ}\) or \(\overrightarrow{PQ}\) or \(\mathbf{PQ}\text{.}\)The magnitude of vector \(\textbf{PQ}\) is represented as \(|\textbf{PQ}|\text{.}\) In this case, we refer to \(P\) as initial point and \(Q\) as the terminal point.
Additionally, a vector can also be represented using a single small letter,such as \(\mathbf{a}\) or \(\underset{\sim}{\mathbf{a}}\text{.}\) In Figure 2.8.3 below, we can represent a vector from a point \(\text{P}\) to point \(Q\) as \(\overrightarrow{PQ} = \mathbf{a} = \underset{\sim}{\mathbf{a}}\)
Figure 2.8.3.
Similarly, if the direction of the vector is reversed, from point \(\text{Q}\) to point \(\text{P}\) the vector is represented as \(\overrightarrow{QP} = \mathbf{-a} = -\underset{\sim}{\mathbf{a}}\)

Subsection 2.8.3 Equivalent Vectors

Activity 2.8.3.

Work in groups
What you require: Graph paper,ruler
(a)
Draw the \(x\) and \(y\) axis on the graph paper.
(b)
Plot the points \(A(0,4), B(3,4), C(0,2)\) and \(D(3,2)\text{.}\)
(c)
Draw a line to connect point \(A\) and \(B,\) add an arrow pointing to point \(B.\)
(d)
Draw a line to connect point \(C\) and \(D,\) add an arrow pointing to point \(D.\)
(e)
What similarities does vector \(\textbf{AB}\) and vector \(\textbf{CD}\) have in common?
(f)
What is the name given to vectors \(\textbf{AB}\) and \(\textbf{CD}\text{?}\)
(g)
Discuss and share your findings with the rest of the class.
\(\textbf{Key Takeaway}\)
Two or more vectors are said to be equivalent if they satisfy the following conditions:
  1. They have same magnitude.
  2. They point in the same direction.
Figure 2.8.4.
In Figure 2.8.4 above, \(\textbf{MN} = \textbf{PQ}\) since they have same direction and equal magnitude.

Example 2.8.5.

Using Figure 2.8.6 below, determine whether vector \(\mathbf{AB}\) and \(\mathbf{DC}\) are equivalent.
Figure 2.8.6.

Solution.

Vector \(\mathbf{AB} \text{ and } \mathbf{DC}\) are equivalent because they have the same magnitude, \(|\mathbf{AB}| = |\mathbf{DC}| \text{,}\) and they point in the same direction

Exercises Exercises

1.
Is it possible for two vectors to have the same direction but not to be equivalent? Explain your answer.
2.
In Figure 2.8.7, identify pairs of equivalent vectors and non-equivalent vectors.
Figure 2.8.7.
3.
Draw two vectors that have the same magnitude and direction but start at different points.

Subsection 2.8.4 Addition of Vectors

Activity 2.8.4.

Work in groups
What you require: Graph paper,ruler
(a)
Draw the \(x\) and \(y\) axis on the graph paper.
(b)
Draw vector \(\textbf{AB}\) from point \(A(0,0)\) to point \(B(2,2)\text{.}\)
(c)
Draw vector \(\textbf{BC}\) from point \(B(2,2)\) to point \(C(5,2)\text{.}\)
(d)
Count the number of units moved horizontally (along the \(x\) axis) from the starting point \(A\) to the final point \(C\text{.}\)
(e)
Similarly, count the number of units moved vertically (along the \(y\) axis) from point \(A\) to point \(C\text{.}\)
(f)
Write the resultant displacement in coordinate form \(\begin{pmatrix} x \\ y \end{pmatrix}\text{,}\) where \(x\) represents displacement along the \(x\) axis and \(y\) represents displacement along the \(y\) axis.
(g)
Discuss and share your findings with the rest of the class.
\(\textbf{Key Takeaway}\)
Consider a displacement from point \(P\) to point \(Q\text{,}\) followed by another displacement from point \(Q\) to point \(N\text{.}\) The total resultant displacement from \(P\) to \(N\) is obtained by adding the two vectors sequentially.
This can be expressed as:
\begin{align*} \mathbf{PN} \amp = \mathbf{PQ} + \mathbf{QN} \end{align*}
where:
  1. \(\mathbf{PQ} = \mathbf{r}\) represents the first displacement.
  2. \(\mathbf{QN} = \mathbf{s}\) represents the second displacement.
Therefore, \(\mathbf{PN} = \mathbf{r} + \mathbf{s}\)
Figure 2.8.8.
We can also re-arrange the two vectors and add them together as shown in Figure 2.8.9 below.
Figure 2.8.9.

Example 2.8.10.

In Figure 2.8.11 below, find vector \(\mathbf{AD}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\text{.}\)
Figure 2.8.11.

Solution.

\begin{align*} \overrightarrow{AD} \amp = \overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CD}\\ \amp = \mathbf{a} + \mathbf{b} + \mathbf{c} \end{align*}
Thus, \(\overrightarrow{AD} = \mathbf{a} + \mathbf{b} + \mathbf{c} \)

Exercises Exercises

1.
\(PQNM\) is a square with vectors \(\mathbf{PQ}\) and \(\mathbf{PM}\) given as \(\mathbf{a} \text{ and } \mathbf{b}\) respectively, as shown in Figure 2.8.12 . Express the \(\mathbf{PN}\) and \(\mathbf{MQ}\) vectors in terms of \(\mathbf{a}\) and \(\mathbf{b}\)
Figure 2.8.12.
2.
Use Figure 2.8.13 to answer the questions below:
Figure 2.8.13.
  1. List pairs of equal vectors.
  2. Name pairs of vectors with equal magnitude but opposite directions.
  3. illustrate the sums of the following vectors graphically:
    1. \(\displaystyle \mathbf{e} + \mathbf{d}\)
    2. \(\displaystyle \mathbf{f} + \mathbf{g}\)
    3. \(\displaystyle \mathbf{b} + \mathbf{d}\)
    4. \(\displaystyle \mathbf{g} + \mathbf{h}\)
3.
Given the vectors \(\mathbf{a} = \binom{2}{3}\) and \(\mathbf{b} = \binom{4}{-1}\text{,}\) find \(\mathbf{a} + \mathbf{b}\) and illustrate the solution graphically.

Subsection 2.8.5 Multiplication of Vectors by a Scalar

Activity 2.8.5.

Work in groups
(a)
On a graph paper, draw the \(x\) axis and \(y\) axis.
(b)
Draw a directed line passing through point \(A(0,2)\) and \(B(2,2)\text{.}\)
(c)
From point \(B(2,2)\text{,}\) draw directed line to point \(C(4,2)\text{.}\)
(d)
Determine the coordinate representations of \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\text{..}\)
(e)
How does \(\overrightarrow{AB}\) relate to \(\overrightarrow{AC}\text{?}\)
(f)
Discuss and share your findings with the rest of the class.
\(\textbf{Key Takeaway}\)
Positive Scalar
In Figure 2.8.14, the vector \(\mathbf{PQ}\) is represented as \(\mathbf{a}\text{.}\) When we multiply \(\mathbf{a}\) by a positive scalar, say \(2\text{,}\) the length of the vector doubles, making it \(\mathbf{2a}\) as shown in Figure 2.8.15. The direction of the vector remains unchanged, but its magnitude increases.
Figure 2.8.14.
In Figure 2.8.15, the vector \(\mathbf{PN}\) is given by: \(\mathbf{PN} = \mathbf{a} + \mathbf{a} = 2\mathbf{a}\text{.}\) This means \(\mathbf{PN}\) has the same direction as \(\mathbf{PQ}\text{,}\) but its magnitude twice that of \(\mathbf{PQ}.\)
Figure 2.8.15.
Negative Scalar
Consider the vector \(\mathbf{AB}\text{,}\)denoted as \(\mathbf{a}\text{,}\) in Figure 2.8.16 The vector points to the right and has a magnitude of \(\mathbf{a}\text{.}\)
Figure 2.8.16.
In Figure 2.8.17, the vector \(\mathbf{AB}\) is obtained by multiplying \(\mathbf{AB}\) by \(-2\text{,}\) giving:
\begin{equation*} \mathbf{AC} = \mathbf{-2} \times \mathbf{a} = \mathbf{-2a} \end{equation*}
This means that \(\mathbf{AC}\) has twice the magnitude of \(\mathbf{AB}\text{,}\) but its direction is reversed.
Figure 2.8.17.
Multiplying a vector by a negative scalar reverses its direction, making it point in the opposite direction.
Zero Scalar
When a vector \(\mathbf{a}\text{,}\) as shown in Figure 2.8.18, is multiplied by \(0\text{,}\) its magnitude becomes \(0\text{,}\) resulting in a zero vector.
Figure 2.8.18.
\begin{equation*} \mathbf{a} \cdot 0 = 0 \end{equation*}

Example 2.8.19.

Given the vectors \(u = 2p + 5q\) and \(v = p - 3q\text{,}\) express \(3u + 2v \) in terms of \(p\) and \(q:\)

Solution.

Substitute the given expressions for u and v.
\begin{align*} 3u + 2v \amp = 3(2p + 5q) + 2(p - 3q) \\ \amp = 6p + 15q + 2p - 6q \end{align*}
Combine like terms (terms with p and terms with q).
\begin{align*} 3u + 2v \amp = (6p + 2p) + (15q - 6q) \\ \amp = 8p + 9q \end{align*}

Exercises Exercises

1.
Simplify the following:
  1. \(\displaystyle 5x + 3y - z + 2(3x - z) + (8x - 6y)\)
  2. \(\displaystyle (\mathbf{a} - \mathbf{b}) + (c - \mathbf{a}) + (\mathbf{b} - c)\)
  3. \(\displaystyle 4m - 2n + 5(k - m) + 2(m + n)\)
2.
Given that \(x = 3m - n\) and \(y = n +4m\text{,}\) express the following vectors in terms of \(m\) and \(n:\)
  1. \(\displaystyle 3x\)
  2. \(\displaystyle \frac{2}{3} y\)
  3. \(\displaystyle 6x - 9y\)
  4. \(\displaystyle 3y - x\)
3.
A pentagon \(ABCDE\) with \(\overrightarrow{AB} = m\text{,}\) \(\overrightarrow{BC} = n\text{,}\) and \(\overrightarrow{CD} = k\text{.}\) Express the following vectors in terms of \(m\text{,}\) \(n\text{,}\) and \(k:\)
  1. \(\displaystyle \overrightarrow{AC}\)
  2. \(\displaystyle \overrightarrow{AD}\)
  3. \(\displaystyle \overrightarrow{AE}\)

Subsection 2.8.6 Column Vectors

Activity 2.8.6.

Work in groups
(a)
Draw the \(x\) axis and \(y\) axis on the graph paper.
(b)
Plot point \(A(1,1)\) and point \(B(5,3)\) on the graph.
(c)
Draw a directed line from point A to point B.
(d)
Represent vector AB in terms of its components as \(\begin{pmatrix} x \\ y \end{pmatrix}\) where \(x\) is the horizontal displacement and \(y\) is the vertical displacement.
(e)
Discuss and share your findings with the rest of the class.
\(\textbf{Key Takeaway}\)
A vector expressed in the form of \(\begin{pmatrix} a \\ b \end{pmatrix}\text{,}\) where \(a\) is the horizontal displacement along the \(x\) axis and \(\mathbf{b}\) is the vertical displacement along the \(y\) axis is known as a column vector.
Figure 2.8.20.
The vector \(\textbf{OP}\) illustrates a displacement from the origin \(O(0,0)\) to the point \(P(4,5)\text{.}\) This consist of a horizontal displacement of \(4\) units along the \(x\) axis and a vertical displacement of \(5\) units in the \(y\) axis.

Example 2.8.21.

Given that: \(\mathbf{a} = \binom{1}{4}\) and \(\mathbf{b} = \binom{5}{3}\text{.}\) Find \(\mathbf{a} + \mathbf{b}\) and illustrate the solution graphically.

Solution.

To determine \(\mathbf{a} + \mathbf{b}\text{,}\) we calculate the total displacement in both the \(x\) and \(y\) directions:
Horizontal displacement is \(1 + 5 = 6.\)
Vertical displacement is \(4 + 3 = 7.\)
Graphical Representation
Begin at the point \((1,0)\) on the grid, move \(1\) unit horizontally to the right and move \(4\) units vertically upwards and mark it as end point. Draw a directed line connecting the two points as shown in the Figure 2.8.22.
From the point \((4,0)\) on the grid, move \(5\) units horizontally to the right parallel to the \(x\) axis, and move \(3\) units vertically up and mark it as end point. Draw another directed line to join the two points.
Now, to find the resultant vector \(\mathbf{a} + \mathbf{b}\text{,}\) join the initial point \((1,0)\) with the final point \((7,7)\) and count the total displacements in the \(x\) and \(y\) directions.
Figure 2.8.22.
Therefore, \(\mathbf{a} + \mathbf{b} = \binom{1}{4} + \binom{5}{3}\)
\(\qquad \qquad \qquad = \binom{6}{7}\)

Example 2.8.23.

If \(\mathbf{a} = \binom{4}{7} \) and \(\mathbf{b} = \binom{3}{5}\text{,}\) find \(2\mathbf{a} + 5\mathbf{b}\text{.}\)

Solution.

To determine \(2\mathbf{a} + 5\mathbf{b}\text{,}\) we multiply vector \(\mathbf{a}\) by \(2\) and vector \(\mathbf{b}\) by \(5\) and finally we add the resulting vectors.
\begin{align*} 2 \mathbf{a} \amp = 2 \binom{4}{7} = \binom{2 \times 4}{2 \times 7} = \binom{8}{14}\\ \amp \\ 5 \mathbf{b} \amp = 5\binom{3}{5} = \binom{5 \times 3}{5 \times 7} = \binom{15}{25} \end{align*}
Therefore, \(2\mathbf{a} + 5\mathbf{b} = \binom{8}{14} + \binom{15}{25} = \binom{8+15}{14+25} = \binom{23}{39}\)

Exercises Exercises

1.
If \(a = \left( \begin{matrix} 2 \\ 5 \end{matrix}\right)\text{,}\) \(b = \left( \begin{matrix} 4 \\ -1 \end{matrix}\right)\) and \(c = \left( \begin{matrix} -7 \\ 3 \end{matrix}\right)\text{,}\) find:
  1. \(\displaystyle 5a + 7c\)
  2. \(\displaystyle -3c + 4b\)
  3. \(\displaystyle \frac{2}{5}a - 2b\)
  4. \(\displaystyle 7c - 2a\)
  5. \(\displaystyle \frac{1}{4}a + \frac{3}{5}b - \frac{1}{2}c\)
  6. \(\displaystyle 5a + 3c - b\)
2.
Given the column vectors \(\mathbf{a} = \binom{3}{-2}\) and \(\mathbf{b} = \binom{-1}{4}\text{,}\) find the following:
  1. \(\displaystyle \mathbf{a} + \mathbf{b}\)
  2. \(\displaystyle 2\mathbf{a} - 3\mathbf{b}\)

Subsection 2.8.7 Position Vectors

Activity 2.8.7.

Work in groups
What you require: Graph paper
(a)
Draw the \(x\) and \(y\) axis on the graph paper as shown below.
(b)
Plot the following points \(A(1,1),B(3,5),C(2,1),D(4,-3) \) on the graph.
(c)
Draw a directed line from \(A\) to \(B\) to represent \(\overrightarrow{AB}\text{.}\)
(d)
Draw another directed line from \(D\) to \(C\) to represent \(\overrightarrow{CD}\text{.}\)
(e)
Determine the position vector of \(B\) relative to point \(A\text{.}\)
(f)
Determine the position vector of \(C\) relative to point \(D\text{.}\)
(g)
Discuss and share your findings with the rest of the class.
\(\textbf{Key Takeaway}\)
In Figure 2.8.24 below, points \(A(2,3)\) and \(B(5,1)\) are located in the plane relative to origin point O in the plane.
The position vector of \(A\) is \(\textbf{OA} = \begin{pmatrix} 2 - 0 \\ 3 - 0 \end{pmatrix} = \binom{2}{3}\)
The position vector of \(B\) is \(\textbf{OB} = \begin{pmatrix} 5 - 0 \\ 1 - 0 \end{pmatrix} = \binom{5}{1}\)
Similarly, for point \(A\) in the plane its position vector \(\textbf{OA}\) is denoted by \(\mathbf{a}\text{.}\) Also for point \(B\) in the plane it’s position vector \(\textbf{OB}\) is denoted by \(\mathbf{b}\text{.}\)
Figure 2.8.24.

Example 2.8.25.

Find the position vector of \(A\) and \(B\text{.}\)

Solution.

The position vector of A is \(\textbf{OA } = \begin{pmatrix} 6 - 0 \\ -2 - 0\end{pmatrix} = \begin{pmatrix} 6 \\ -2 \end{pmatrix}\)
Similarly the position vector of B is \(\textbf{OB }= \begin{pmatrix} 4 - 0 \\ -4 - 0\end{pmatrix} = \begin{pmatrix} 4 \\ -4 \end{pmatrix}\)

Exercises Exercises

1.
Draw the following position vector on a graph paper:
\(a = \left( \begin{matrix} 10 \\ -2 \end{matrix}\right)\)
\(b = \left( \begin{matrix} 2 \\ 5 \end{matrix}\right)\)
\(c = \left( \begin{matrix} -3 \\ 8 \end{matrix}\right)\)
\(d = \left( \begin{matrix} -11 \\ 6 \end{matrix}\right)\)
\(e = \left( \begin{matrix} 4 \\ 4 \end{matrix}\right)\)
\(f = \left( \begin{matrix} -1 \\ 12 \end{matrix}\right)\)
\(g = \left( \begin{matrix} 5 \\ 3 \end{matrix}\right)\)
\(h = \left( \begin{matrix} 0 \\ 13 \end{matrix}\right)\)
2.
Use Figure 2.8.26 below to write the position vectors of points \(M,E,D,A\text{.}\)
Figure 2.8.26.

Subsection 2.8.8 Magnitude of a Vector

Activity 2.8.8.

Work in groups
What you require: Graph paper
(a)
Draw the \(x\) axis and \(y\) axis on the graph paper.
(b)
Mark the coordinate \((0,0)\) as the initial point \(O\text{.}\)
(c)
From Point \(O\text{,}\) move \(3\) units to the right along the \(x\) axis and \(4\) units upward in the \(y\) axis. Mark this new position as Point \(A\text{.}\)
(d)
Draw a directed line from point \(O\) to point \(A\) to represent \(\overrightarrow{OA}\text{.}\)
(e)
Use a ruler to measure the length of \(\overrightarrow{OA}\text{.}\)
(f)
Analyze the relationship between the \(x\) displacement, \(y\) displacement, and the length of \(\overrightarrow{OA}\text{.}\)
(g)
Discuss and share your findings with your classmates in the class.
\(\textbf{Key Takeaway}\)
The magnitude of \(\overrightarrow{AB}\) in Figure 2.8.27 can be denoted as \(|\mathbf{AB}|\text{.}\) The magnitude of \(\overrightarrow{AB}\) represents the distance between point \(\text{A}\) and point \(\text{B}\text{.}\)
We can represent the components of \(\overrightarrow{AB}\) as \(\begin{pmatrix} x \\ y \end{pmatrix}\text{,}\) where \(x\) represents the horizontal displacement and \(y\) represents the vertical displacement.
We determine the magnitude of \(\overrightarrow{AB}\) by applying Pythagorean theorem as shown below.
\begin{equation*} |\mathbf{AB}| = \sqrt{x^2 + y^2} \end{equation*}
Figure 2.8.27.
The magnitude of a vector is always positive since \(x\) and \(y\) components are squared, resulting in \(x^2\) and \(y^2\text{,}\) both of which are non-negative.

Example 2.8.28.

Determine the magnitude of \(\overrightarrow{AB}\) as shown in the Figure 2.8.29 below.
Figure 2.8.29.

Solution.

To find the magnitude of \(\overrightarrow{AB}\text{,}\) we apply pythagora’s theorem to find the length \(\textbf{AB}\text{.}\)
\begin{align*} |\textbf{AB}| \amp = \sqrt{\text{(AC)}^2 + \text{(CB)}^2 } \\ \amp = \sqrt{7^2 + 24^2} \\ \amp = \sqrt{49 + 576}\\ \amp = \sqrt{625}\\ \amp = 25 \end{align*}
Hence, the magnitude of \(\overrightarrow{AB}\) represented as \(|\textbf{AB}|\) is \(25.\)

Example 2.8.30.

Given that \(a=\binom{2}{4}\text{,}\) \(b=\binom{-2}{2.5}\text{,}\) \(c=\binom{6}{-4}\) and \(r=a+2b-c\text{.}\) Find \(|r|\)

Solution.

\(r=a+2b-c\text{.}\)
Substituting the values of \(a\text{,}\) \(b\) and \(c\) into the equation;
\(r=\binom{2}{4}+ 2\binom{-2}{2.5}-\binom{6}{-4}\)
\(r=\binom{2}{4}+ \binom{-4}{5}-\binom{6}{-4}\)
\(r=\binom{2+(-4)-6}{4+5-(-4)}\)
\(r=\binom{-8}{13}\)
\(|r|=\sqrt{(-8)^2+13^2}=\sqrt{64+169}=\sqrt{233}\)
\(r=15.26\)

Exercises Exercises

1.
Find the magnitude of each of the following vectors:
  1. \(\displaystyle \left( \begin{matrix} -6 \\ 8 \end{matrix}\right)\)
  2. \(\displaystyle \left( \begin{matrix} 8 \\ 15 \end{matrix}\right)\)
  3. \(\displaystyle \left( \begin{matrix} -5 \\ 12 \end{matrix}\right)\)
  4. \(\displaystyle \left( \begin{matrix} 3 \\ 7 \end{matrix}\right)\)

Subsection 2.8.9 Midpoint of a Vector

Activity 2.8.9.

Work in groups
What you require: Graph paper
(a)
Draw the \(x\) and \(y\) axis on the graph paper.
(b)
Choose any starting point on the graph and label it as Point \(A\text{.}\) write down its coordinates.
(c)
From Point \(A\text{,}\) move \(6\) units to the right parallel to the \(x\) axis and mark this new location as Point \(B\text{.}\) Write down its coordinates.
(d)
Draw a directed line from point \(A\) to Point \(B\) to represent \(\overrightarrow{AB}\text{.}\)
(e)
Find the midpoint of \(\overrightarrow{AB}\) and label it as Point \(M\text{.}\)
(f)
Identify the coordinates of Point \(M\text{.}\)
(g)
Think of a way to determine coordinates of Point \(M\) without manually counting the units.
(h)
Discuss and share your findings with the rest of the class.
\(\textbf{Key Takeaway}\)
Consider the cordinates of point \(P\) given as \((x_1,y_1)\) and point \(N\) given as \((x_2,y_2)\) and \(M\) is the midpoint of \(\mathbf{PN}\) as shown in figure below.
Figure 2.8.31.
\begin{align*} \mathbf{OM} \amp = \mathbf{OP} + \mathbf{PM} \\ \amp = \mathbf{a} + \frac{1}{2} \mathbf{PN}\\ \amp = \mathbf{a} + \frac{1}{2} (\mathbf{b} - \mathbf{a})\\ \amp = \mathbf{a} + \frac{1}{2}\mathbf{b} - \frac{1}{2}\mathbf{a}\\ \amp = \mathbf{a} - \frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}\\ \amp = \frac{\mathbf{a} + \mathbf{b}}{2} \end{align*}
But \(\mathbf{a} = \begin{pmatrix} x_1 \\ y_1 \end{pmatrix}\) and \(\mathbf{b} = \begin{pmatrix} x_2 \\ y_2 \end{pmatrix}\)
Thus, midpoint \(\mathbf{M} = \left( \frac{ x_1 + x_2 }{ 2}, \frac{ y_1 + y_2 }{ 2}\right)\)

Example 2.8.32.

Find the coordinates of the midpoint \(M\) of \(\textbf{AB}\) given the following points \(A(6,1), \,\, B(4,3)\text{.}\)

Solution.

To determine the midpoint \(M\) of \(\overrightarrow{AB}\) , we apply the midpoint formulae as follows.
\begin{align*} \text{Midpoint } \amp = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) \\ \amp = \left(\frac{6 + 4}{2}, \frac{1+3}{2}\right)\\ \amp = \left(\frac{10}{2}, \frac{4}{2}\right)\\ \amp = \left(5, 2\right) \end{align*}
Thus, the coordinates of the midpoint \(M\) is \((5,2)\)

Exercises Exercises

1.
Find the coordinates of the midpoint of \(\mathbf{PQ}\) in each of the following cases:
  1. \(\displaystyle P\, (-5,6), Q\, (3,-4)\)
  2. \(\displaystyle P\,(1,−3), Q\,(5,7)\)
  3. \(\displaystyle P\, (-2,4), Q\, (6,0)\)
  4. \(\displaystyle P\, (a,b), Q\, (c,d)\)
2.
In the figure below, \(ABC\) is a triangle where \(M\) and \(N\) are midpoints of \(\overrightarrow{AC}\text{,}\) \(\overrightarrow{AB}\) respectively.
Triangle \(ABC\) has vertices at points \(A(2,2)\text{,}\) \(B(6,2)\) and \(C(4,6)\text{.}\) Determine the coordinates of \(M\) and \(N\text{.}\)
3.
Consider triangle \(PQR\) where \(\overrightarrow{PQ} = \mathbf{u}\) and \(\overrightarrow{PR} = \mathbf{v}.\) Point \(S\) is located on \(PR\) such that \(PS:SR = 2:3\text{.}\) Point \(T\) lies on \(QR\) with \(QT:TR = 3:2\text{,}\) and point \(U\) is on \(PQ\) such that \(PU:UQ = 4:1\text{.}\) Determine the following vectors in terms of \(\mathbf{u}\) and \(\mathbf{v}\text{:}\)
  1. \(\displaystyle \overrightarrow{PS} \)
  2. \(\displaystyle \overrightarrow{PT} \)
  3. \(\displaystyle \overrightarrow{PU} \)
  4. \(\displaystyle \overrightarrow{ST} \)
  5. \(\displaystyle \overrightarrow{TU} \)

Subsection 2.8.10 Translation Vector

Activity 2.8.10.

Work in groups
What you require: Graph paper
(a)
Draw the \(x\) axis and \(y\) axis on the graph paper.
(b)
Plot the triangle with vertices \(A(-3,1)\text{,}\) \(B(-1,1)\text{,}\) and \(C(-2,3)\text{.}\)
(c)
Translate each point by moving \(2\) units to the right parallel to the \(x\) axis and \(3\) units up in the \(y\) axis. Label the new points as \(A', B',\) and \(C'\text{.}\)
(d)
Draw the new triangle \(A'B'C'\) on the graph.
(e)
Use dotted lines to connect each original point to its corresponding translated point \((A \text{ to } A', B \text{ to } B', C \text{ to } C')\text{,}\) add arrows to indicate the direction.
(f)
Observe and describe any similarities between triangle \(ABC\) and triangle \(A'B'C'.\)
(g)
Analyze the distance each point moved.
(h)
Discuss and share your findings with your classmates in the class.
\(\textbf{Key Takeaway}\)
A square \(\mathbf{ABCD}\) undergoes a translation when each of its vertices (\(\mathbf{A}\text{,}\)\(\mathbf{B}\text{,}\)\(\mathbf{C}\) and \(\mathbf{D}\)) is moved the same distance and in the same direction. A translation vector, denoted by \(\mathbf{T}\text{,}\) describes this movement.
Using \(\mathbf{T}\) to represent a translation, the notation \(\mathbf{T}(\mathbf{P})\) indicates the application of the translation \(\mathbf{T}\) on \(\mathbf{P}\text{.}\) In Figure 2.8.33, shows \(\mathbf{A'B'C'D'}\) is the image of \(\mathbf{ABCD}\) under a translation.
Figure 2.8.33.

Example 2.8.34.

Triangle \(\mathbf{ABC}\) has vertices \(\mathbf{A}(1,3)\text{,}\) \(\mathbf{B}(3,0)\) and \(\mathbf{C}(4,4)\text{.}\)The triangle undergoes a translation \(\mathbf{T}\) defined by the vector \(\begin{pmatrix} 4 \\ 3 \end{pmatrix}\text{.}\)
  1. Determine the coordinates of the translated vertices \(\mathbf{A'}\text{,}\) \(\mathbf{B'}\) , and \(\mathbf{C'}\) ’.
  2. Draw the triangle \(\mathbf{ABC}\) and its image under \(\mathbf{T}\text{.}\)

Solution.

To find the coordinates of the translated vertices, we apply the translation \(\mathbf{T}\) to each original vertex.
\begin{align*} \mathbf{OA'} \amp = \mathbf{OA} + \mathbf{T}\\ \mathbf{OB'} \amp = \mathbf{OB} + \mathbf{T}\\ \mathbf{OC'} \amp = \mathbf{OC} + \mathbf{T} \end{align*}
\begin{align*} \mathbf{OA'} \amp = \binom{1}{3} + \binom{4}{3} = \binom{1+4}{3+3} = \binom{5}{6} \\ \mathbf{OB'} \amp = \binom{3}{0} + \binom{4}{3} = \binom{3+4}{0+3} = \binom{7}{3} \\ \mathbf{OC'} \amp = \binom{4}{4} + \binom{4}{3} = \binom{4+4}{4+3} = \binom{8}{7} \end{align*}
Figure 2.8.35.

Exercises Exercises

1.
Draw triangle \(\mathbf{XYZ}\) with vertices \(\mathbf{X(1, 4)}\text{,}\) \(\mathbf{Y(6, 2)}\text{,}\) and \(\mathbf{Z(5, 3)}\text{.}\) On the same axes, plot \(\mathbf{X'Y'Z'}\text{,}\) the image of triangle \(\mathbf{XYZ}\) under a translation given by \(\binom{4}{9}.\)
2.
The following points have been translated using the given vectors. Determine their original positions:
  1. \(\displaystyle (4,-1); \, \, \binom{2}{3}\)
  2. \(\displaystyle (0,-3); \, \, \binom{1}{3}\)
  3. \(\displaystyle (-3,8); \, \, \binom{-2}{7}\)
  4. \(\displaystyle (11,-5); \,\, \binom{5}{-1}\)
  5. \(\displaystyle (-12,5); \,\, \binom{3}{-10}\)
  6. \(\displaystyle (2,-7); \,\, \binom{-10}{15}\)
3.
A point \(\mathbf{P(5, -3)}\) is mapped to a new position after a translation. If the new coordinates are \(\mathbf{(9, 1)}\text{,}\) determine the translation vector used.
4.
A point \(\mathbf{M(1, -4)}\) undergoes a translation by \(\binom{3}{5}.\) Determine the coordinates of \(\mathbf{M'}\text{,}\) the transformed point. If \(\mathbf{M'}\) is then translated by \(\binom{-4}{2},\) find the final position \(\mathbf{M''}\text{.}\) What is the single translation vector that maps \(\mathbf{M'}\) to \(\mathbf{M''}\) directly?
5.
Translate each of the following points using the given vector:
  1. \(\displaystyle (10,1); \, \, \binom{-11}{2}\)
  2. \(\displaystyle (-2,-5); \, \, \binom{6}{14}\)
  3. \(\displaystyle (3,-15); \, \, \binom{-16}{11}\)
  4. \(\displaystyle (-11,4); \, \, \binom{-15}{10}\)
  5. \(\displaystyle (1,10); \, \, \binom{7}{2}\)
  6. \(\displaystyle (4,-9); \, \, \binom{5}{-8}\)
  7. \(\displaystyle (-2,13); \, \, \binom{-3}{1}\)
  8. \(\displaystyle (-6,5); \, \, \binom{-1}{2}\)

Reading Questions Further Exercises

1.
Given that \(a=\) \((-3,2)\) , \(b=\) \((6,-4)\) and \(c=\) \((5,-15) \)and that \(q=\) \(2a\)+ \(\frac{1}{2}b\) +\(\frac{1}{5}c\text{.}\) Express \(q\) as a column vector and hence calculate its magnitude \(|q|\) correct to two decimal places
2.
If \(P\text{,}\) \(Q\) and \(R\) are the points \((2, - 4)\text{,}\) \((4, 0)\) and \((1, 6)\) respectively, use the vector method to find the coordinates of point \(S\) given that \(PQRS\) is a parallelogram.
3.
The figure below shows a triangle of vectors in which \(OS: SP=\) \(1:3\text{,}\) \(PR:RQ=\) \(2:1\) and \(X\) is the midpoint of \(OR\text{.}\)
Given that \(OP=\) \(p\) and \(OQ=\) \(q\text{,}\) express the following vectors in terms of \(p\) and \(q\)
(i) \(PQ\)
(ii) \(QR\)
(iii) \(OR\)
(iv) \(QT\)
4.
Given that \(3p-2q=\binom{10}{5}\) and \(p+2q=\binom{-7}{15}\) find;
  1. \(p\) and \(q\)
  2. \(\displaystyle |2p+3q|\)
5.
In triangle \(OAB\text{,}\) \(M\) and \(N\) are points on \(OA\) and \(OB\) respectively, such that \(OM:OA=2:5\) and \(ON:OB=2:3\text{.}\) \(AN\) and \(BM\) intersect at \(T\text{.}\) Given that \(OA=a\) and \(OB=b\text{.}\)Express in terms of \(a\) and \(b\text{:}\)
(a). \(AB\)
(b) \(BM\)
(C) \(AN\)
6.
Find the scalars \(p\) and \(q\)
\(p\binom{8}{6}+q \binom{-6}{4}=\binom{10}{16}\)
7.
A point \(M\) divides \(PQ\) in the ratio \(3:5\text{.}\) Given that \(P\) is \((-6,8)\) and \(Q\) is \((4,-2)\text{.}\) Find the coordinates of \(M\)
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