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.
