Question

if a = (-3, 1) and b = (1, 0), what is the angle between the two vectors? round

Explanation:

Step1: Recall dot - product formula

The dot - product of two vectors $\vec{a}=(a_1,a_2)$ and $\vec{b}=(b_1,b_2)$ is $\vec{a}\cdot\vec{b}=a_1b_1 + a_2b_2$, and $\vec{a}\cdot\vec{b}=\vert\vec{a}\vert\vert\vec{b}\vert\cos\theta$, where $\theta$ is the angle between the two vectors, and $\vert\vec{a}\vert=\sqrt{a_1^{2}+a_2^{2}}$, $\vert\vec{b}\vert=\sqrt{b_1^{2}+b_2^{2}}$. Given $\vec{a}=(-3,1)$ and $\vec{b}=(1,0)$.
First, calculate the dot - product:
$\vec{a}\cdot\vec{b}=(-3)\times1 + 1\times0=-3$.

Step2: Calculate the magnitudes of the vectors

$\vert\vec{a}\vert=\sqrt{(-3)^{2}+1^{2}}=\sqrt{9 + 1}=\sqrt{10}$, $\vert\vec{b}\vert=\sqrt{1^{2}+0^{2}} = 1$.

Step3: Find the cosine of the angle

Since $\vec{a}\cdot\vec{b}=\vert\vec{a}\vert\vert\vec{b}\vert\cos\theta$, then $\cos\theta=\frac{\vec{a}\cdot\vec{b}}{\vert\vec{a}\vert\vert\vec{b}\vert}$. Substitute the values we found: $\cos\theta=\frac{-3}{\sqrt{10}\times1}=-\frac{3}{\sqrt{10}}$.

Step4: Find the angle

$\theta=\arccos(-\frac{3}{\sqrt{10}})\approx 108.43^{\circ}$ (rounded to two decimal places).

Answer:

$\theta\approx108.43^{\circ}$