Answer:
[tex]\theta=\text{ 145.3\degree}[/tex]
Step-by-step explanation:
The angle between two vectors is represented by the following equation:
[tex]\cos \theta=\frac{\vec{u}\cdot\vec{v}}{\lvert\vec{u}\rvert\lvert\vec{v}\rvert}[/tex]
Notice that it involves a trigonometric function, the dot product of two vectors, and the magnitude of two vectors.
Then, let's determine the dot product of the two vectors:
[tex]\begin{gathered} \vec{v}\cdot\vec{w}=-4\cdot2+-3\cdot6 \\ \vec{v}\cdot\vec{w}=-26 \end{gathered}[/tex]
Now, calculate the magnitudes of the vectors:
[tex]\begin{gathered} \lvert\vec{v}\rvert=\sqrt[]{(-4)^2+(-3)^2}=5 \\ \lvert\vec{w}\rvert=\sqrt[]{(2)^2+(6)^2}=2\sqrt[]{10} \end{gathered}[/tex]
Now, substitute into the equation to find the angle:
[tex]\begin{gathered} \cos \theta=\frac{-26}{5\cdot2\sqrt[]{10}} \\ \theta=\cos ^{-1}(\frac{-26}{5\cdot2\sqrt[]{10}}) \\ \theta=\text{ 145.3\degree} \end{gathered}[/tex]
