Question

What is the magnitude of the position vector whose terminal point is (-2, 4)?

Answer 1

Answer:

  2√5

Step-by-step explanation:

The Pythagorean theorem tells you how to find the distance from the origin.

  d = √((-2)² +4²) = √20 = 2√5

The vector's magnitude is 2√5 ≈ 4.47214.

Answer 2

Answer:

The magnitude of the position is $|x|=\sqrt{20}$

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,

$|x|=\sqrt{a^2+b^2}$

Let point x=(-2,4)

$|x|=\sqrt{(-2)^2+(4)^2}$

$|x|=\sqrt{4+16}$

$|x|=\sqrt{20}$

Therefore, The magnitude of the position is $|x|=\sqrt{20}$