Sweet! That wasn't very difficult, was it?

Given a vector's magnitude and its angle with the x-axis, we know how to calculate its x- and its y-components. How about the

Yes! If \(\vec{A} = A_x \hat{i} + A_y \hat{j}\), then, the magnitude is just \(\sqrt{A_x^2 + A_y^2}\). The angle is calculated by taking the inverse tan of \(\frac{A_y}{A_x}\)

**reverse**? Given the components, can we calculate the magnitude and angle?Yes! If \(\vec{A} = A_x \hat{i} + A_y \hat{j}\), then, the magnitude is just \(\sqrt{A_x^2 + A_y^2}\). The angle is calculated by taking the inverse tan of \(\frac{A_y}{A_x}\)