Question

A vector has an x-component of -25.0 units and a y-component of 40.0 units. Find the magnitude and direction of the vector.

Answer 1

Given:

$$a_x=-25.0,\quad a_y=40.0$$

The magnitude of the vector $\bf a$

$$a=\sqrt{a_x^2+a_y^2}=\sqrt{(-25.0)^2+40.0^2}=47.2$$

The angle between vector $\bf a$ and positive direction of x-axis

$$\theta=\tan^{-1}\frac{a_y}{a_x}=\pi-\tan^{-1}\frac{40.0}{25.0}=122^{\circ}$$