Question

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

2sqrt17
4
5
6

Answer 1

`√(2\cdot 17)`

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|=√(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|=√((-5)^2+(3)^2)=√(25+9)=√(34)=√(2\cdot 17)`