Answer:
115.6° (1 d.p.)
Step-by-step explanation:
To find the angle between two vectors:
- Create a triangle with the vectors as two sides and the included angle θ between them.
- Find the magnitude of each vector (the length of each side of the triangle).
- Use the cosine rule to find the angle θ.
**Please see attached for the triangle diagram**
Given vectors:
[tex]\textbf{u}=-4\textbf{i}-3\textbf{j}[/tex]
[tex]\textbf{v}=-\textbf{i}+5\textbf{j}[/tex]
Use Pythagoras Theorem to find the magnitude of each vector:
[tex]\implies |\textbf{u}|=\sqrt{(-4)^2+(-3)^2}=5[/tex]
[tex]\implies |\textbf{v}|=\sqrt{(-1)^2+5^2}=\sqrt{26}[/tex]
[tex]\overrightarrow{\text{UV}}=\textbf{v}-\textbf{u}=(-\textbf{i}+5\textbf{j})-(-4\textbf{i}-3\textbf{j})=3\textbf{i}+8\textbf{j}[/tex]
[tex]|\overrightarrow{\text{UV}}|=\sqrt{3^2+8^2}=\sqrt{73}[/tex]
Cosine Rule (for finding angles)
[tex]\sf \cos(C)=\dfrac{a^2+b^2-c^2}{2ab}[/tex]
where:
- C = angle
- a and b = sides adjacent the angle
- c = side opposite the angle
Find angle θ using the cosine rule:
[tex]\implies \cos(\theta)=\dfrac{|\textbf{u}|^2+|\textbf{v}|^2-|\overrightarrow{\text{UV}}|^2}{2|\textbf{u}||\textbf{v}|}[/tex]
[tex]\implies \cos(\theta)=\dfrac{5^2+\left(\sqrt{26}\right)^2-\left(\sqrt{73}\right)^2}{2(5)\left(\sqrt{26}\right)}[/tex]
[tex]\implies \cos(\theta)=\dfrac{-22}{10\sqrt{26}}[/tex]
[tex]\implies \theta=\cos^{-1}\left(\dfrac{-22}{10\sqrt{26}}\right)[/tex]
[tex]\implies \theta=115.5599652...^{\circ}[/tex]
Therefore, the angle between the vectors is 115.6° (1 d.p.).
