Representing Vectors

Subsection 2.9.2 Representing Vectors

Vectors are typically represented using arrows. The length of the arrow represents the magnitude of the vector, and the direction of the arrow shows the direction of the vector.

🔗

Subsection 2.9.2.1 Representing Vectors Geometrically

Curriculum Alignment

Strand 🔗: 2.0 Measurements and Geometry
🔗
Sub-Strand 🔗: 2.9 Vectors 1
🔗

Specific Learning Outcome 🔗: Add vectors in different situations
🔗

🔗

Teacher Resource 2.9.9.

To assist your teaching, we have prepared lesson resources, aligned with this textbook and the CBC. The Lesson Plan links to syllabus learning outcomes and provides suggest time allocations. The Step-by-Step Guide provides more detailed guidance on how to teach the content, including suggested questions to ask learners, and possible answers.

Learner Experience 2.9.2.

Work in Groups

🔗

Materials: ruler, graph paper, pencil.

🔗

(a)

Draw vector \(\overrightarrow{AB}\) from \(A(0,0) to B(4,2)\text{.}\) Measure its magnitude using the distance formula.

🔗

(b)

Draw another vector \(\overrightarrow{BC}\) from \(B(4,2)\) to \(C(6,5)\text{.}\) Add \(\overrightarrow{AB}\) and \(\overrightarrow{BC}\) .

🔗

(c)

Draw two vectors from the origin: \(\mathbf{a} = (3,1)\) and \(\mathbf{b} = (2,4)\text{.}\) Construct the parallelogram to determine \(\mathbf{a} + \mathbf{b}\text{.}\)

🔗

(d)

Compare with other groups and discuss the results.

🔗

Key Takeaway 2.9.10.

We can represent a vector by a line drawn between two points as shown in Figure 2.9.11 below:

🔗

Figure 2.9.11.

🔗

In Figure 2.9.11, the direction of the vector is indicated by an arrow pointing to the right, while its magnitude is determined by the length of \(PQ\text{.}\)

🔗

A vector \(PQ\) can be denoted as \(\underset{\sim}{PQ}\) or \(\overrightarrow{PQ}\text{.}\)The magnitude of vector \(PQ\) is represented as \(|PQ|\text{.}\) 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 \(\underset{\sim}{\mathbf{a}}\text{.}\)

🔗

Example 2.9.12.

Find the magnitude of vector \(\overrightarrow{PQ}\) where \(P(1,2)\) and \(Q(5,5)\text{.}\)

🔗

Solution.

Step 1: Determine the components.

🔗

\(\overrightarrow{PQ} = (5-1, 5-2) = (4,3)\)

🔗

Step 2: Use the magnitude formula.

🔗

\(|\overrightarrow{PQ}| = \sqrt{4^2 + 3^2}\)

🔗

\(= \sqrt{16 + 9} = \sqrt{25} = 5\)

🔗

Example 2.9.13.

Add vectors \(\mathbf{a} = (2,3)\) and \(\mathbf{b} = (4,1)\text{.}\)

🔗

Solution.

Add corresponding components.

🔗

\(\mathbf{a} + \mathbf{b} = (2+4, 3+1)\)

🔗

\(= (6,4)\)

🔗

Therefore, the resultant vector is \((6,4)\text{.}\)

🔗

Example 2.9.14.

Determine \(\mathbf{a} - \mathbf{b}\) if \(\mathbf{a} = (5,2)\) and \(\mathbf{b} = (1,4)\text{.}\)

🔗

Solution.

Subtract corresponding components.

🔗

\(\mathbf{a} - \mathbf{b} = (5-1, 2-4)\)

🔗

\(= (4,-2)\)

🔗

Therefore, the resulting vector is \((4,-2)\text{.}\)

🔗

Exercises Exercises

1.

Find the magnitude of vector \((3,4)\text{.}\)

🔗

Answer.

\(\sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\)

🔗

2.

Add vectors \((1,2)\) and \((3,5)\text{.}\)

🔗

Answer.

\((1+3, 2+5) = (4,7)\)

🔗

3.

Given \(A(0,0), B(2,1)\text{,}\) and \(C(5,4)\text{,}\) determine \(\overrightarrow{AB} + \overrightarrow{BC}\text{.}\)

🔗

Answer.

\(\overrightarrow{AB} = (2,1)\)

🔗

\(\overrightarrow{BC} = (3,3)\)

🔗

\((2,1) + (3,3) = (5,4)\)

🔗

Therefore, \(\overrightarrow{AB} + \overrightarrow{BC} = (5,4)\text{.}\)

🔗

4.

If \(\mathbf{a} = (6,-2)\) and \(\mathbf{b} = (4,3)\text{,}\) find \(2\mathbf{a}\text{.}\)

🔗

Answer.

\(2\mathbf{a} = (12,-4)\)

🔗

Subsubsection 2.9.2.1 Column Vectors

Curriculum Alignment

Strand 🔗: 2.0 Measurements and Geometry
🔗
Sub-Strand 🔗: 2.9 Vectors 1
🔗

Specific Learning Outcome 🔗: Multiply vectors by scalars
🔗

🔗

Teacher Resource 2.9.15.

Learner Experience 2.9.3.

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.

🔗

Key Takeaway 2.9.16.

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 2.9.17.

🔗

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 2.9.18.

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 2.9.19.

🔗

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 2.9.19.

🔗

Therefore, \(\mathbf{a} + \mathbf{b} = \binom{1}{4} + \binom{5}{3}\)

🔗

\(\qquad \qquad \qquad = \binom{6}{7}\)

🔗

Example 2.9.20.

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:

🔗

\(\displaystyle 5a + 7c\)
🔗

🔗
\(\displaystyle -3c + 4b\)
🔗

🔗
\(\displaystyle \frac{2}{5}a - 2b\)
🔗

🔗
\(\displaystyle 7c - 2a\)
🔗

🔗
\(\displaystyle \frac{1}{4}a + \frac{3}{5}b - \frac{1}{2}c\)
🔗

🔗
\(\displaystyle 5a + 3c - b\)
🔗

🔗

🔗

Answer.

\(\displaystyle \binom{-39}{46}\)
🔗

🔗
\(\displaystyle \binom{37}{-13}\)
🔗

🔗
\(\displaystyle \binom{\frac{-36}{5}}{4}\)
🔗

🔗
\(\displaystyle \binom{-53}{11}\)
🔗

🔗
\(\displaystyle \binom{\frac{32}{5}}{\frac{23}{20}}\)
🔗

🔗
\(\displaystyle \binom{-15}{35}\)
🔗

🔗

🔗

2.

Given the column vectors \(\mathbf{a} = \binom{3}{-2}\) and \(\mathbf{b} = \binom{-1}{4}\text{,}\) find the following:

🔗

\(\displaystyle \mathbf{a} + \mathbf{b}\)
🔗

🔗
\(\displaystyle 2\mathbf{a} - 3\mathbf{b}\)
🔗

🔗

🔗

Answer.

\(\displaystyle \binom{2}{2}\)
🔗

🔗
\(\displaystyle \binom{9}{-16}\)
🔗

🔗

🔗

Checkpoint 2.9.21.

🔗

Checkpoint 2.9.22.

🔗

Subsubsection 2.9.2.2 Position Vectors

Curriculum Alignment

Strand 🔗: 2.0 Measurements and Geometry
🔗
Sub-Strand 🔗: 2.9 Vectors 1
🔗

Specific Learning Outcome 🔗: Determine translation vector as a transformation
🔗

🔗

Teacher Resource 2.9.23.

Learner Experience 2.9.4.

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.

🔗

Key Takeaway 2.9.24.

In Figure 2.9.25 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 2.9.25.

🔗

Example 2.9.26.

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 \(\textbf{OB }= \begin{pmatrix} 4 - 0 \\ -4 - 0\end{pmatrix} = \begin{pmatrix} 4 \\ -4 \end{pmatrix}\)

🔗

Example 2.9.27.

The points \(A(1,2)\text{,}\) \(B(7,4)\) and \(C(3,10)\) lie in the Cartesian plane.

🔗

Find the coordinates of point \(D\) which divides \(BC\) internally in the ratio \(2:1\text{.}\)
🔗

🔗
Find the coordinates of point \(E\text{,}\) the midpoint of \(AD\text{.}\)
🔗

🔗
Show that \(A\text{,}\) \(E\) and the centroid of triangle \(ABC\) are collinear.
🔗

🔗
Find the area of triangle \(AEC\text{.}\)
🔗

🔗

🔗

Solution.

Since \(D\) divides \(BC\) internally in the ratio \(2:1\text{,}\)
🔗

\(D = \frac{2C + 1B}{3}\)
🔗

\begin{align*} D = \amp \frac{2(3,10) + (7,4)}{3}\\ = \amp \frac{(6,20) + (7,4)}{3}\\ = \amp \frac{(13,24)}{3} \end{align*}

🔗

\(D\left(\frac{13}{3},8\right)\)
🔗

🔗
Since \(E\) is the midpoint of \(AD\text{,}\)
🔗

\(E = \frac{A + D}{2}\)
🔗

\begin{align*} E = \amp \frac{\left(1,2\right) + \left(\frac{13}{3},8\right)}{2}\\ = \amp \frac{\left(\frac{16}{3},10\right)}{2} \end{align*}

🔗

\(E\left(\frac{8}{3},5\right)\)
🔗

🔗
The centroid \(G\) of triangle \(ABC\) is
🔗

\begin{align*} G = \amp \frac{A + B + C}{3}\\ = \amp \frac{(1,2) + (7,4) + (3,10)}{3}\\ = \amp \frac{(11,16)}{3} \end{align*}

🔗

\(G\left(\frac{11}{3}, \frac{16}{3}\right)\)
🔗

Now consider vectors:
🔗

\(\vec{AE} = \left(\frac{8}{3}-1, 5-2\right) = \left(\frac{5}{3},3\right)\)
🔗

\(\vec{AG} = \left(\frac{11}{3}-1, \frac{16}{3}-2\right) = \left(\frac{8}{3}, \frac{10}{3}\right)\)
🔗

\(\vec{AG} \ne x \vec{AE}\)
🔗

Therefore, the vectors are not scalar multiples, hence \(A\text{,}\) \(E\) and \(G\) are not collinear.
🔗

🔗
Area of triangle \(AEC\text{:}\)
🔗

Use Surveyor’s Formula:
🔗

\begin{align*} \text{Area} =\amp \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \\ = \amp \frac{1}{2} \left|1 \times -\frac{1}{3} + \frac{8}{3} \times \frac{10}{3} + \frac{11}{3} \right| \times -3 \\ = \amp \frac{1}{2} \left| -\frac{1}{3} + \frac{80}{9} -11 \right|\\ = \amp \frac{1}{2} \times \frac{22}{9}\\ = \amp \frac{11}{9} \end{align*}

🔗

Therefore, the area of triangle \(AEC\) is \(\frac{11}{9}\) square units.
🔗

🔗