Question

find the magnitude of \\(\vec{b} - \vec{a}\\). express your answer in meters. the diagram shows vectors: \\(\vec{a}\\) (8.00 m) along negative x-axis, \\(\vec{b}\\) (15.0 m) at 30.0° from positive y-axis towards x-axis, \\(\vec{d}\\) (10.0 m) at 53.0° from positive y-axis, \\(\vec{c}\\) (12.0 m) at 25.0° from negative y-axis (or positive x-axis?); input field for \\(|\vec{b} - \vec{a}|\\) in meters.

Explanation:

Step1: Find components of \(\vec{A}\)

\(\vec{A}\) is along the negative x - axis. So, \(A_x=- 8.00\space m\), \(A_y = 0\space m\).

Step2: Find components of \(\vec{B}\)

The angle of \(\vec{B}\) with the positive y - axis is \(30.0^{\circ}\), so with the positive x - axis, the angle \(\theta=90^{\circ}- 30.0^{\circ}=60.0^{\circ}\) below the positive x - axis.
\(B_x = 15.0\cos(60.0^{\circ})=15.0\times0.5 = 7.5\space m\)
\(B_y=-15.0\sin(60.0^{\circ})=-15.0\times\frac{\sqrt{3}}{2}\approx - 12.99\space m\)

Step3: Find components of \(\vec{B}-\vec{A}\)

\((\vec{B}-\vec{A})_x=B_x - A_x=7.5-(-8.00)=15.5\space m\)
\((\vec{B}-\vec{A})_y=B_y - A_y=- 12.99 - 0=-12.99\space m\)

Step4: Calculate the magnitude of \(\vec{B}-\vec{A}\)

Using the formula \(|\vec{B}-\vec{A}|=\sqrt{(\vec{B}-\vec{A})_x^{2}+(\vec{B}-\vec{A})_y^{2}}\)
\(|\vec{B}-\vec{A}|=\sqrt{(15.5)^{2}+(- 12.99)^{2}}=\sqrt{240.25 + 168.74}=\sqrt{408.99}\approx20.2\space m\)

Answer:

\(20.2\space m\) (The value may vary slightly depending on the precision of \(\sin(60^{\circ})\) and \(\cos(60^{\circ})\) used. If we use more precise values, for example, \(\sin(60^{\circ})=\frac{\sqrt{3}}{2}\approx0.8660\), then \(B_y=-15\times0.8660 = - 12.99\), \(B_x = 7.5\), \((\vec{B}-\vec{A})_x=15.5\), \((\vec{B}-\vec{A})_y=-12.99\), and \(|\vec{B}-\vec{A}|=\sqrt{15.5^{2}+12.99^{2}}=\sqrt{240.25 + 168.76}=\sqrt{409.01}\approx20.22\space m\), which can be rounded to \(20.2\space m\) or \(20.22\space m\))