Question

Vector vector u equals vector PQ has initial point P (5, 16) and terminal point Q (8, 4). Vector vector v equals vector RS has initial point R (28, 7) and terminal point S (10, 16).Write u and v in trigonometric form.

Answer 1

u = 12.37(0.242, -0.970)

v = 20.12(-0.89, 0.45)

Step-by-step explanation

We want to write u and v in trigonometric form.

1. for u

Determine the component form of u

`\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}`

Determine the magnitude of the vector u

`\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}`

Determine the angle theta

`\begin{gathered} \tan \text{ }\theta\text{ = }(y)/(x) \\ \tan \text{ }\theta\text{ = -}(12)/(3) \\ \tan \text{ }\theta\text{ = -4} \\ \theta=tan^(-1)(-4) \\ \theta=-76^0 \end{gathered}`

But this is in the wrong quadrant

so,

`\theta=-76+360=284^0`
`\begin{gathered} \sin (284)\text{ = -}0.970 \\ \cos (284)\text{ = 0.242} \end{gathered}`

Hence, writing u in trigonometric form:

u = 12.37(0.242, -0.970).

2. for v

Determine the component form of v

`\begin{gathered} \text{component form of v = } \\ \text{ = <(x}_2-x_1).(y_2-y_1)> \\ \text{ = <-18, 9>} \end{gathered}`

Determine the magnitude of the vector v

`\begin{gathered} \Vert v\Vert\text{ =}\sqrt[]{(-18)^2+(9)^2} \\ \text{ = }\sqrt[]{405} \\ \text{ = 20.12} \end{gathered}`

Determine the angle theta

`\begin{gathered} \tan \text{ }\theta\text{ = -}(9)/(18) \\ \theta\text{ = }\tan ^(-1)(-0.5) \\ \theta=-27^0 \end{gathered}`

But this is in the wrong quadrant

so,

`\theta=-27+180=153^0`
`\begin{gathered} \sin (153)\text{ = 0.454} \\ \cos (153)\text{ = -0.89} \end{gathered}`

Hence, writing u in trigonometric form:

v = 20.12(-0.89, 0.45)