Answer:
The magnitude of the position is [tex]|x|=\sqrt{20}[/tex]
Step-by-step explanation:
Given : Vector whose terminal point is (-2, 4).
To find : What is the magnitude of the position vector?
Solution :
We have given, terminal point (-2,4)
The magnitude of the point x(a,b) is given by,
[tex]|x|=\sqrt{a^2+b^2}[/tex]
Let point x=(-2,4)
[tex]|x|=\sqrt{(-2)^2+(4)^2}[/tex]
[tex]|x|=\sqrt{4+16}[/tex]
[tex]|x|=\sqrt{20}[/tex]
Therefore, The magnitude of the position is [tex]|x|=\sqrt{20}[/tex]
