Respuesta :
ANSWER
u = 12.37(0.242, -0.970)
v = 20.12(-0.89, 0.45)
EXPLANATION
We want to write u and v in trigonometric form.
1. for u
Determine the component form of u
[tex]\begin{gathered} \text{component form of u = } \\ \text{ = <(x}_2-x_1).(y_2-y_1)> \\ =\text{ <(8-5), (4-16)>} \\ \text{ = <3, -12>} \end{gathered}[/tex]Determine the magnitude of the vector u
[tex]\begin{gathered} \mleft\Vert u\mleft\Vert\text{ =}\sqrt[]{3^2+(-12)^2}\mright?\mright? \\ \text{ = }\sqrt[]{9+144} \\ \text{ = }\sqrt[]{153} \\ \mleft\Vert u\mleft\Vert\text{ = 12.37}\mright?\mright? \end{gathered}[/tex]Determine the angle theta
[tex]\begin{gathered} \tan \text{ }\theta\text{ = }\frac{y}{x} \\ \tan \text{ }\theta\text{ = -}\frac{12}{3} \\ \tan \text{ }\theta\text{ = -4} \\ \theta=tan^{-1}(-4) \\ \theta=-76^0 \end{gathered}[/tex]But this is in the wrong quadrant
so,
[tex]\theta=-76+360=284^0[/tex][tex]\begin{gathered} \sin (284)\text{ = -}0.970 \\ \cos (284)\text{ = 0.242} \end{gathered}[/tex]Hence, writing u in trigonometric form:
u = 12.37(0.242, -0.970).
2. for v
Determine the component form of v
[tex]\begin{gathered} \text{component form of v = } \\ \text{ = <(x}_2-x_1).(y_2-y_1)> \\ \text{ = <-18, 9>} \end{gathered}[/tex]Determine the magnitude of the vector v
[tex]\begin{gathered} \Vert v\Vert\text{ =}\sqrt[]{(-18)^2+(9)^2} \\ \text{ = }\sqrt[]{405} \\ \text{ = 20.12} \end{gathered}[/tex]Determine the angle theta
[tex]\begin{gathered} \tan \text{ }\theta\text{ = -}\frac{9}{18} \\ \theta\text{ = }\tan ^{-1}(-0.5) \\ \theta=-27^0 \end{gathered}[/tex]But this is in the wrong quadrant
so,
[tex]\theta=-27+180=153^0[/tex][tex]\begin{gathered} \sin (153)\text{ = 0.454} \\ \cos (153)\text{ = -0.89} \end{gathered}[/tex]Hence, writing u in trigonometric form:
v = 20.12(-0.89, 0.45)
