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

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.

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danielmaduroh danielmaduroh · Answer 2 · 2018-09-11T02:03:33+02:00

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

[tex]\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[/tex]

[tex]\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[/tex]

Therefore:

[tex]\vec{A}=(38.85, 20.65)m \\ \\ \vec{B}=(-14.81, 21.97)m[/tex]

So:

[tex]\vec{B}-\vec{A}=(-14.81, 21.97)-(38.85, 20.65)=(-53.66,1.32)[/tex]

Finally, the magnitude is:

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