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Vectors u = −2(cos 30°i + sin30°j), v = 6(cos 225°i + sin225°j), and w = 8(cos 120°i + sin120°j) are given. Use exact values when evaluating sine and cosine.

Part A: Convert the vectors to component form and find −7(u • v). Show every step of your work. (4 points)

Part B: Convert the vectors to component form and use the dot product to determine if u and w are parallel, orthogonal, or neither. Justify your answer. (6 points)

Vectors U 2cos 30i Sin30j V 6cos 225i Sin225j And W 8cos 120i Sin120j Are Given Use Exact Values When Evaluating Sine And Cosine Part A Convert The Vectors To C class=

Sagot :

Vectors

Component Form

The component form of a vector looks like this:

<x, y>  or (x, y),

where x and y are the horizontal and vertical components that make up the vector's orientation.

To convert a vector in polar form ||v||(cosΘ, sinΘ) to component form, evaluate the product in each component!

[tex]\dotfill[/tex]

Dot Product

The dot product

                                 [tex]\bold v \cdot \bold u=||\bold v||\:||\bold u||cos\theta[/tex],

where ||v|| and ||u|| are the magnitude of each vector and theta is the angle (in radians or degrees) between them.

                       

If the dot product is

  • equal to zero means that the vectors are perpendicular to each other
  • equal to the product of the vectors' magnitudes which means that they are parallel to each other
  • equal to a value other than the ones listed above it means it is neither parallel nor perpendicular

Why?

If two vectors are perpendicular to each other, they have a 90-degree or [tex]\dfrac{\pi}{2}[/tex] between them. Plugging that into the dot product:

                               [tex]\bold v \cdot \bold u = ||\bold v||\: ||\bold u||cos\left(\dfrac{\pi}{2}\right) =||\bold v||\:||\bold u||(0)\\\bold v \cdot \bold u = 0[/tex].

Similarly, if two vectors are parallel, they have no angle (0) in between. Plugging that into the dot product:

                               [tex]\bold v \cdot \bold u = ||\bold v||\: ||\bold u||cos\left(0\right) =||\bold v||\:||\bold u||(1)\\\\\bold v \cdot \bold u =||\bold v||\:||\bold u||[/tex].

[tex]\hrulefill[/tex]

Solving the Problem

Part A

We evaluate each component in the given polar form of each vector to find their component form!
               [tex]\bold u = -2(cos(30^\circ)i,\:sin(30^\circ)j)=-2\left(\dfrac{\sqrt{3} }{2} ,\:\dfrac{1}{2} \right)\\\boxed{\bold u = (-\sqrt3, \:1)}[/tex]

               [tex]\bold v = 6(cos(225^\circ)i,\:sin(225^\circ)j)=6\left(-\dfrac{\sqrt{2} }{2} ,\:-\dfrac{\sqrt2}{2} \right)\\\boxed{\bold v = (-3\sqrt2, \:-3\sqrt2)}[/tex]

We can solve for [tex]-7(\bold u \cdot \bold v)[/tex].

                       [tex]-7(\bold u \cdot \bold v) = -7 [(-2 \times 6)cos(225-30)^\circ]\\\\-7(\bold u \cdot \bold v)=-7[-12cos(195^\circ)]\\\\\boxed{-7(\bold u \cdot \bold v)=-81.14}[/tex]

[tex]\dotfill[/tex]

Part B

We've converted vector u into component form but not vector w.

                   [tex]\bold w = 8(cos(120^\circ)i,\:sin(120^\circ)j)=8\left(-\dfrac{1 }{2} ,\:\dfrac{\sqrt3}{2} \right)\\\boxed{\bold w = (-4, \:4\sqrt3)}[/tex]

The dot product of u and w is

                      [tex]\bold w \cdot \bold u = (8\times -2)cos(120-30)^\circ =-16(0)\\\boxed{\bold w\cdot \bold u =0}[/tex].

This means that vectors u and w are perpendicular to each other.

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