Question

(d) what is the magnitude and direction of b - c?

Explanation:

Step1: Recall magnitude formula

The magnitude of a vector with components $x$ and $y$ is given by $|\vec{V}|=\sqrt{x^{2}+y^{2}}$. For the vector $\vec{B}-\vec{C}$, if the $x -$ component is $- 4$ and the $y -$ component is $8$, then $|\vec{B}-\vec{C}|=\sqrt{(-4)^{2}+8^{2}}=\sqrt{16 + 64}=\sqrt{80}=4\sqrt{5}$.

Step2: Recall direction formula

The direction $\theta$ of a two - dimensional vector with components $x$ and $y$ is given by $\theta=\tan^{-1}(\frac{y}{x})$. Here, $x=-4$ and $y = 8$, so $\theta=\tan^{-1}(\frac{8}{-4})=\tan^{-1}(-2)$. Since $x<0$ and $y>0$, the vector lies in the second quadrant, and $\theta = 180^{\circ}+\tan^{-1}(-2)\approx180^{\circ}- 63.43^{\circ}=116.57^{\circ}$ (in degrees) or $\theta=\pi+\tan^{-1}(-2)$ (in radians).

Answer:

Magnitude: $4\sqrt{5}$; Direction: $\theta=\tan^{-1}(-2)+180^{\circ}$ (in degrees) or $\theta=\pi+\tan^{-1}(-2)$ (in radians)