Question

question suppose that $vec{u}=langle - 9,1
angle$ and $vec{v}=langle 5,-7
angle$. what is the angle $\theta$ between $vec{u}$ and $vec{v}$, rounded to the nearest degree? provide your answer below: $\theta=square^{circ}$

Explanation:

Step1: Recall dot - product formula

The dot - product formula is $\vec{u}\cdot\vec{v}=\vert\vec{u}\vert\vert\vec{v}\vert\cos\theta$, where $\vec{u}\cdot\vec{v}=u_1v_1 + u_2v_2$, $\vert\vec{u}\vert=\sqrt{u_1^{2}+u_2^{2}}$, $\vert\vec{v}\vert=\sqrt{v_1^{2}+v_2^{2}}$, and $\theta$ is the angle between $\vec{u}$ and $\vec{v}$. First, calculate $\vec{u}\cdot\vec{v}$:
$\vec{u}\cdot\vec{v}=(-9)\times5+1\times(-7)=-45 - 7=-52$.

Step2: Calculate magnitudes of vectors

Calculate $\vert\vec{u}\vert$: $\vert\vec{u}\vert=\sqrt{(-9)^{2}+1^{2}}=\sqrt{81 + 1}=\sqrt{82}$.
Calculate $\vert\vec{v}\vert$: $\vert\vec{v}\vert=\sqrt{5^{2}+(-7)^{2}}=\sqrt{25 + 49}=\sqrt{74}$.

Step3: Solve for $\cos\theta$

From $\vec{u}\cdot\vec{v}=\vert\vec{u}\vert\vert\vec{v}\vert\cos\theta$, we have $\cos\theta=\frac{\vec{u}\cdot\vec{v}}{\vert\vec{u}\vert\vert\vec{v}\vert}$.
Substitute the values: $\cos\theta=\frac{-52}{\sqrt{82}\times\sqrt{74}}=\frac{-52}{\sqrt{82\times74}}=\frac{-52}{\sqrt{6068}}\approx\frac{-52}{77.9}$.
$\cos\theta\approx - 0.6675$.

Step4: Find the angle $\theta$

$\theta=\arccos(-0.6675)$. Using a calculator, $\theta\approx131.8^{\circ}\approx153^{\circ}$ (rounded to the nearest degree).

Answer:

$153^{\circ}$