Respuesta :
Answer:
[tex]v=<\sqrt{32}cos(319.11^{\circ}),\sqrt{32}sin(319.11^{\circ})>[/tex] or [tex]v=\sqrt{32}cos(319.11^{\circ})i+\sqrt{32}sin(319.11^{\circ})j[/tex]
Step-by-step explanation:
Find the magnitude of [tex]<2\sqrt{6},-2\sqrt{2}>[/tex]:
[tex]||v||=\sqrt{(2\sqrt{6})^2+(-2\sqrt{2})^2}[/tex]
[tex]||v||=\sqrt{24+8}[/tex]
[tex]||v||=\sqrt{32}[/tex]
Find the direction angle of [tex]<2\sqrt{6},-2\sqrt{2}>[/tex]:
[tex]\alpha=tan^{-1}|\frac{-2\sqrt{2}}{2\sqrt{6}}|[/tex]
[tex]\alpha=tan^{-1}|-\frac{\sqrt{3}}{2}|[/tex]
[tex]\alpha=tan^{-1}(\frac{\sqrt{3}}{2})[/tex]
[tex]\alpha=40.89^{\circ}[/tex]
Since [tex]\alpha=40.89^{\circ}[/tex] is our reference angle, we determine our direction angle [tex]\theta[/tex] by verifying our angle is in Quadrant IV, which is where the vector is located. Therefore, [tex]\theta=360^{\circ}-\alpha=360^{\circ}-40.89^{\circ}=319.11^{\circ}[/tex]
To represent a vector's magnitude and direction in trigonometric form, we use the form [tex]v=<||v||cos\theta,||v||sin\theta>[/tex] or [tex]v=||v||cos\theta i+||v||sin\theta j[/tex].
In conclusion, the trigonometric form of the vector is:
[tex]v=<\sqrt{32}cos(319.11^{\circ}),\sqrt{32}sin(319.11^{\circ})>[/tex] or
[tex]v=\sqrt{32}cos(319.11^{\circ})i+\sqrt{32}sin(319.11^{\circ})j[/tex]
