Curriculum Alignment
- Strand
- 2.0 Measurements and Geometry
- Sub-Strand
- 2.9 Vectors 1
Subsection 2.9.3 Magnitudes of Vectors
Teacher Resource 2.9.31.
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 2.9.5.
(a)
(b)
(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)
(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.
Key Takeaway 2.9.32.
The magnitude of \(\overrightarrow{AB}\) in Figure 2.9.33 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*}
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 2.9.34.
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*}
Example 2.9.36.
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.
\(r=a+2b-c\text{.}\)
\(r=\binom{2}{4}+ 2\binom{-2}{2.5}-\binom{6}{-4}\)
\(r=\binom{2}{4}+ \binom{-4}{5}-\binom{6}{-4}\)
\(r=\binom{2+(-4)-6}{4+5-(-4)}\)
\(r=\binom{-8}{13}\)
\(|r|=\sqrt{(-8)^2+13^2}=\sqrt{64+169}=\sqrt{233}\)
\(r=15.26\)
Exercises Exercises
1.
Find the magnitude of each of the following vectors:
-
\(\displaystyle \left( \begin{matrix} -6 \\ 8 \end{matrix}\right)\)
-
\(\displaystyle \left( \begin{matrix} 8 \\ 15 \end{matrix}\right)\)
-
\(\displaystyle \left( \begin{matrix} -5 \\ 12 \end{matrix}\right)\)
-
\(\displaystyle \left( \begin{matrix} 3 \\ 7 \end{matrix}\right)\)
Answer.
-
\(\displaystyle \sqrt{(-6)^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10\)
-
\(\displaystyle \sqrt{8^2 + 15^2} = \sqrt{64 + 225} = \sqrt{289} = 17\)
-
\(\displaystyle \sqrt{(-5)^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13\)
-
\(\displaystyle \sqrt{3^2 + 7^2} = \sqrt{9 + 49} = \sqrt{58} \approx 7.62\)
