Question

if $v_1=langle - 8,9
angle$ and $v_2=langle7,0
angle$, 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: $\vec{v_1}\cdot\vec{v_2}=(-8)\times7+9\times0=-56$.

Step2: Calculate magnitudes of vectors

The magnitude of a vector $\vec{v}=(x,y)$ is $\vert\vec{v}\vert=\sqrt{x^{2}+y^{2}}$. For $\vec{v_1}=(-8,9)$, $\vert\vec{v_1}\vert=\sqrt{(-8)^{2}+9^{2}}=\sqrt{64 + 81}=\sqrt{145}$. For $\vec{v_2}=(7,0)$, $\vert\vec{v_2}\vert=\sqrt{7^{2}+0^{2}} = 7$.

Step3: Solve for 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: $\cos\theta=\frac{-56}{7\sqrt{145}}=\frac{-8}{\sqrt{145}}\approx\frac{-8}{12.04159}\approx - 0.6643$. Then $\theta=\cos^{-1}(-0.6643)\approx131.69^{\circ}$.

Answer:

$131.69$