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

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

`\bullet \ cos(28^(\circ))=(Adjacent)/(Hypotenuse)=(A_(x))/(44) \\ \\ \therefore A_(x)=44cos(28^(\circ))=38.85m \\ \\ \\ \bullet \ sin(28^(\circ))=(Opposite)/(Hypotenuse)=(A_(y))/(44) \\ \\ \therefore A_(y)=44sin(28^(\circ))=20.65m`

`\bullet \ cos(56^(\circ))=(Adjacent)/(Hypotenuse)=(-B_(x))/(26.5) \\ \\ \therefore B_(x)=-26.5cos(56^(\circ))=-14.81m \\ \\ \\ \bullet \ sin(56^(\circ))=(Opposite)/(Hypotenuse)=(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|=√((-53.66)^2+(1.32)^2)=53.67m}`