MATH SOLVE

4 months ago

Q:
# Vector u has a magnitude of 7 units and a direction angle of 330°. Vector v has magnitude of 8 units and a direction angle of 30°. What is the direction angle of their vector sum?

Accepted Solution

A:

keeping in mind that x = rcos(θ) and y = rsin(θ).

we know the magnitude "r" of U and V, as well as their angle θ, so let's get them in standard position form.

[tex]\bf u= \begin{cases} x=7cos(330^o)\\ \qquad 7\cdot \frac{\sqrt{3}}{2}\\ \qquad \frac{7\sqrt{3}}{2}\\ y=7sin(330^o)\\ \qquad 7\cdot -\frac{1}{2}\\ \qquad -\frac{7}{2} \end{cases}\qquad \qquad v= \begin{cases} x=8cos(30^o)\\ \qquad 8\cdot \frac{\sqrt{3}}{2}\\ \qquad \frac{8\sqrt{3}}{2}\\ y=8sin(30^o)\\ \qquad 8\cdot \frac{1}{2}\\ \qquad 4 \end{cases}[/tex]

[tex]\bf u+v\implies \left( \frac{7\sqrt{3}}{2},-\frac{7}{2} \right)+\left( \frac{8\sqrt{3}}{2},4 \right)\implies \left( \frac{7\sqrt{3}}{2}+\frac{8\sqrt{3}}{2}~~,~~ -\frac{7}{2}+4\right) \\\\\\ \left(\stackrel{a}{\frac{15\sqrt{3}}{2}}~~,~~ \stackrel{b}{\frac{1}{2}}\right)\\\\ -------------------------------[/tex]

[tex]\bf tan(\theta )=\cfrac{b}{a}\implies tan(\theta )=\cfrac{\frac{1}{2}}{\frac{15\sqrt{3}}{2}}\implies tan(\theta )=\cfrac{1}{15\sqrt{3}} \\\\\\ \measuredangle \theta =tan^{-1}\left( \cfrac{1}{15\sqrt{3}} \right)\implies \measuredangle \theta \approx 2.20422750397203^o[/tex]

we know the magnitude "r" of U and V, as well as their angle θ, so let's get them in standard position form.

[tex]\bf u= \begin{cases} x=7cos(330^o)\\ \qquad 7\cdot \frac{\sqrt{3}}{2}\\ \qquad \frac{7\sqrt{3}}{2}\\ y=7sin(330^o)\\ \qquad 7\cdot -\frac{1}{2}\\ \qquad -\frac{7}{2} \end{cases}\qquad \qquad v= \begin{cases} x=8cos(30^o)\\ \qquad 8\cdot \frac{\sqrt{3}}{2}\\ \qquad \frac{8\sqrt{3}}{2}\\ y=8sin(30^o)\\ \qquad 8\cdot \frac{1}{2}\\ \qquad 4 \end{cases}[/tex]

[tex]\bf u+v\implies \left( \frac{7\sqrt{3}}{2},-\frac{7}{2} \right)+\left( \frac{8\sqrt{3}}{2},4 \right)\implies \left( \frac{7\sqrt{3}}{2}+\frac{8\sqrt{3}}{2}~~,~~ -\frac{7}{2}+4\right) \\\\\\ \left(\stackrel{a}{\frac{15\sqrt{3}}{2}}~~,~~ \stackrel{b}{\frac{1}{2}}\right)\\\\ -------------------------------[/tex]

[tex]\bf tan(\theta )=\cfrac{b}{a}\implies tan(\theta )=\cfrac{\frac{1}{2}}{\frac{15\sqrt{3}}{2}}\implies tan(\theta )=\cfrac{1}{15\sqrt{3}} \\\\\\ \measuredangle \theta =tan^{-1}\left( \cfrac{1}{15\sqrt{3}} \right)\implies \measuredangle \theta \approx 2.20422750397203^o[/tex]