Question

if $v_1=langle - 5,6
angle$ and $v_2=langle3,0
angle$, 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.
$\vec{v_1}\cdot\vec{v_2}=(- 5)\times3+6\times0=-15$

Step2: Calculate the magnitudes of the vectors

The magnitude of a vector $\vec{v}=(x,y)$ is $\vert\vec{v}\vert=\sqrt{x^{2}+y^{2}}$.
$\vert\vec{v_1}\vert=\sqrt{(-5)^{2}+6^{2}}=\sqrt{25 + 36}=\sqrt{61}$
$\vert\vec{v_2}\vert=\sqrt{3^{2}+0^{2}} = 3$

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 $\cos\theta=\frac{\vec{v_1}\cdot\vec{v_2}}{\vert\vec{v_1}\vert\vert\vec{v_2}\vert}$.
$\cos\theta=\frac{-15}{3\sqrt{61}}=-\frac{5}{\sqrt{61}}\approx - 0.64$

Step4: Find the angle

$\theta=\arccos(-0.64)\approx130.0^{\circ}$

Answer:

$130.0^{\circ}$