Answer:
Check the graph below for the sketch.
[tex]\theta=135^{\circ}[/tex]
Step-by-step explanation:
1) Let's organize the data from these Latex codes.
[tex]\vec{\mathbf{u}}=\left \langle 4,0 \right \rangle\:and\: \vec{\mathbf{v}}=\left \langle -3,-3 \right \rangle[/tex]
2) Sketching them (Check below). Notice that we are going to use the coordinates of each vector.
3) To find the angle between the vectors, we need to remember the Theorem:
If there is an angle between two vectors a and b, then we can calculate its angle using this relation:
[tex]a*b=\left \| a \right \|\left \| b \right \|*cos\theta[/tex]
3.1) Then we need the Dot product between u and v. The Dot product is going to give us a scalar value for a product between vectors.
We will also need the norm of each vector. The norm will tell us the length of each one.
[tex]\vec{u}*\vec{v}=\vec{u}*\vec{v}=4(-3)+0(-3)=-12\\\left \| u \right \|=\sqrt{4^2+0^2 }= \left \| u \right \|=4, \\\left \| v \right \|=\sqrt{(-3)^2+(-3)^2}, \left \| v \right \|=18=3\sqrt{2} \left\\[/tex]
3.2)
Now, we can plug these pieces of information from 3.1 and 3.2 and find the angle:
[tex]cos\theta=\frac{uv}{\left \| u \right \|\left \| v \right \|}=\frac{4*-3+(0)(-3)}{12\sqrt{2} }=\frac{-12}{ 12\sqrt{2} }\\\theta=cos^{-1}(\frac{-12}{ 12\sqrt{2} })\rightarrow \theta\approx 135^{\circ}[/tex]
