Question

Given the vectors A⃗ and B⃗ shown in the figure ((Figure 1) ), determine the magnitude of B⃗ −A⃗. A is 28 degrees above the positive x-axis and is 44m long, and B is 56 degrees above the negative x-axis and is 26.5m long

Answer 1

To answer this problem, we are given with two vectors. We translate these into angle form such that the mode complex in the calculator fits in the computation. We input 44<28 + 26.5< (180-56). The answer from the calculator is 48.93< 60.59. The magnitude is 48.93 pointing to the direction of 60.59 degrees above the x axis.

Answer 2

This problem is represented in the Figure below. So, we can find the components of each vector as follows:

$\bullet \ cos(28^{\circ})=\frac{Adjacent}{Hypotenuse}=\frac{A_{x}}{44} \\ \\ \therefore A_{x}=44cos(28^{\circ})=38.85m \\ \\ \\ \bullet \ sin(28^{\circ})=\frac{Opposite}{Hypotenuse}=\frac{A_{y}}{44} \\ \\ \therefore A_{y}=44sin(28^{\circ})=20.65m$

$\bullet \ cos(56^{\circ})=\frac{Adjacent}{Hypotenuse}=\frac{-B_{x}}{26.5} \\ \\ \therefore B_{x}=-26.5cos(56^{\circ})=-14.81m \\ \\ \\ \bullet \ sin(56^{\circ})=\frac{Opposite}{Hypotenuse}=\frac{B_{y}}{26.5} \\ \\ \therefore B_{y}=26.5sin(56^{\circ})=21.97m$

Therefore:

$\vec{A}=(38.85, 20.65)m \\ \\ \vec{B}=(-14.81, 21.97)m$

So:

$\vec{B}-\vec{A}=(-14.81, 21.97)-(38.85, 20.65)=(-53.66,1.32)$

Finally, the magnitude is:

$\boxed{\left| \vec{B}-\vec{A}\right|=\sqrt{(-53.66)^2+(1.32)^2}=53.67m}$