Question

Find the magnitude of the vector:​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

$\large\boxed{|\overrightarrow{CD}|=3\sqrt{61}}$

Step-by-step explanation:

The formula of a magnitude of vector:

$|\overrightarrow{AB}|=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

where

$A(x_1,\ y_1),\ B(x_2,\ y_2)$

We have:

$C(10,\ 10),\ D(-8,\ -5)$

Substitute:

$|\overrightarrow{CD}|=\sqrt{(-8-10)^2+(-5-10)^2}\\\\|\overrightarrow{CD}|=\sqrt{(-18)^2+(-15)^2}\\\\|\overrightarrow{CD}|=\sqrt{324+225}\\\\|\overrightarrow{CD}|=\sqrt{549}\\\\|\overrightarrow{CD}|=\sqrt{(9)(61)}\\\\|\overrightarrow{CD}|=\sqrt{9}\cdot\sqrt{61}\\\\|\overrightarrow{CD}|=3\sqrt{61}$