Question

A vector points into the third quadrant, and its x and y components are both negative. If its magnitude is equal to two times the magnitude of its x component, what is the angle between the vector and the positive x axis?

Answer 1

Answer:

$240^{\circ}$

Explanation:

The vector lies in the 3rd quadrant, which means that both its x and y components are negative.

Let's analyze the magnitude now. We know that the magnitude of the vector (let's call it v) is 2 times the magnitude of the x-component:

$|v| = 2 |v_x|$

we know that the magnitude of the x-component is given by

$|v_x | = |v| cos \theta$

where here $\theta$ is measured as angle below the negative x-direction, since we are in the 3rd quadrant. Substituting into the previous equation,

$|v| = 2 |v| cos \theta\\cos \theta = \frac{1}{2}\\\theta = 60^{\circ}$

But this is the angle between the vector and the negative x axis: therefore, the angle between the vector and the positive x axis is

$\theta=60^{\circ} + 180^{\circ}=240^{\circ}$