Answer:
See below
Step-by-step explanation:
3. What are two ways that a vector can be represented?
Considering a vector [tex]\vec{v}[/tex] in some vector space [tex]\mathbb R^n[/tex] we have
[tex]\vec{v} = \langle a,b\rangle[/tex]
This is the component form. I don't like that way. It is probably used in high school, but
[tex]\vec{v} = \begin{pmatrix} a\\ b\\ \end{pmatrix}[/tex]
is preferable because the inner product on [tex]\mathbb R^n[/tex] is defined to be
[tex]$\langle a,b\rangle := \sum_{i = 1}^n a_i b_i$[/tex]
You can also write it using linear form such as [tex]\vec{v} = 2i+2j[/tex]
4.
For this question, I think you meant
vectors
[tex]\vec{u_1} = (-8, 12)[/tex]
[tex]\vec{u_2} = (13, 15)[/tex]
Once
[tex]\cos(\theta)=\dfrac{\vec{u_1} \cdot\vec{u_2}}{||\vec{u_1}||||\vec{u_2}||}[/tex]
Considering that the dot product is
[tex]\vec{u_1}\cdot \vec{u_2} = (-8)\cdot 13 + 12\cdot 15 = -104+180= 76[/tex]
and the norm of [tex]\vec{u_1}[/tex] is [tex]||\vec{u_1}|| = \sqrt{(-8)^2 + 12^2} = \sqrt{64 + 144}= \sqrt{208}[/tex]
and the norm of [tex]\vec{u_2}[/tex] is [tex]||\vec{u_2}|| = \sqrt{13^2 + 15^2} = \sqrt{169 + 225}= \sqrt{394}[/tex]
Thus,
[tex]\cos(\theta)=\dfrac{76}{\sqrt{208} \sqrt{394}} = \dfrac{19}{\sqrt{13}\sqrt{394}}=\dfrac{19}{\sqrt{5122}}[/tex]
[tex]\therefore \theta = \arccos \left(\dfrac{19}{\sqrt{5122}} \right)[/tex]
