Question

if v1 = (-5, 4) and v2 = (9, 0), what is the angle between the two vectors? round to two decimals.

Explanation:

Step1: Recall dot - product formula

The dot - product of two vectors $\vec{v_1}=(x_1,y_1)$ and $\vec{v_2}=(x_2,y_2)$ is $\vec{v_1}\cdot\vec{v_2}=x_1x_2 + y_1y_2$, and $\vec{v_1}\cdot\vec{v_2}=\vert\vec{v_1}\vert\vert\vec{v_2}\vert\cos\theta$, where $\theta$ is the angle between the two vectors. First, calculate the dot - product:
$\vec{v_1}\cdot\vec{v_2}=(- 5)\times9+4\times0=-45$.

Step2: Calculate the magnitudes of the vectors

The magnitude of a vector $\vec{v}=(x,y)$ is $\vert\vec{v}\vert=\sqrt{x^{2}+y^{2}}$.
For $\vec{v_1}=(-5,4)$, $\vert\vec{v_1}\vert=\sqrt{(-5)^{2}+4^{2}}=\sqrt{25 + 16}=\sqrt{41}$.
For $\vec{v_2}=(9,0)$, $\vert\vec{v_2}\vert=\sqrt{9^{2}+0^{2}} = 9$.

Step3: Solve for the cosine of the angle

Since $\vec{v_1}\cdot\vec{v_2}=\vert\vec{v_1}\vert\vert\vec{v_2}\vert\cos\theta$, then $\cos\theta=\frac{\vec{v_1}\cdot\vec{v_2}}{\vert\vec{v_1}\vert\vert\vec{v_2}\vert}$.
Substitute the values: $\cos\theta=\frac{-45}{9\sqrt{41}}=-\frac{5}{\sqrt{41}}\approx - 0.7809$.

Step4: Find the angle

$\theta=\arccos(-0.7809)$. Using a calculator, $\theta\approx141.34^{\circ}$.

Answer:

$141.34^{\circ}$