Question

What is the magnitude of the position vector whose terminal point is (-5, 3)?

Answer 1

Answer:

$\sqrt{2\cdot 17}$

Step-by-step explanation:

The magnitude of a vector is the length of the vector itself.

Given a bi-dimensional vector, the magnitude of the vector is given by:

$|v|=\sqrt{v_x^2+v_y^2}$

where

$v_x$ is the x-component of the vector

$v_y$ is the y-component of the vector

The vector in this problem is

$v=(-5,3)$

Therefore its components are

$v_x=-5\\v_y=3$

And so, the magnitude of the vector is:

$|v|=\sqrt{(-5)^2+(3)^2}=\sqrt{25+9}=\sqrt{34}=\sqrt{2\cdot 17}$