Answers:
- Choice D
- Choice C
- Choice C
- Choice D
- Choice B
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Explanations:
Problem 1
v = 9i - 4j means we move 9 units to the right and 4 units down.
Another way to write this is v = <9, -4>
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Problem 2
R = (-19, 6)
S = (-35, 13)
Subtract x coordinates: xS - xR = -35 - (-19) = -16
Subtract y coordinates: yS - yR = 13 - 6 = 7
Those results form the vector v = -16i + 7j
This vector indicates we move 16 units left and 7 units up when going from point R(-19,6) to point S(-35,13).
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Problem 3
t = -5i + 12j
||t|| = sqrt( (-5)^2 + (12)^2 )
||t|| = 13
theta = arctan(12/(-5))
theta = -67.38
Add on 180 to get to the proper quadrant Q2
-67.38 + 180 = 112.62
Therefore,
[tex]t = 13\bigg\langle \cos(112.62^{\circ}), \ \sin(112.62^{\circ})\bigg\rangle[/tex]
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Problem 4
The x, y components are the coefficients of the linear form.
This allows us to go immediately from <-39,16> to -39i + 16j
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Problem 5
P = (-5,11)
Q = (-16,4)
Subtract x coords: xQ - xP = -16 - (-5) = -11
Subtract y coords: yQ - yP = 4 - 11 = -7
The component form of vector t is <-11, -7> and this vector is in Q3.
r = sqrt(a^2 + b^2)
r = sqrt( (-11)^2 + (-7)^2 )
r = 13.038 is the approximate length of the vector.
theta = arctan(b/a)
theta = arctan(-7/(-11))
theta = 32.471
Add on 180 so we get to Q3
32.471+180 = 212.471
Therefore, we have
[tex]x = r\cos(\theta) = 13.038\cos(212.471^{\circ})\\\\y = r\sin(\theta) = 13.038\sin(212.471^{\circ})\\\\t = xi + yj\\\\t = 13.038\cos(212.471^{\circ})*i + 13.038\sin(212.471^{\circ})*j\\\\[/tex]
in which you could write into this slightly more compact form
[tex]t = 13.038\bigg\langle\cos(212.471^{\circ}),\sin(212.471^{\circ})\bigg\rangle\\\\[/tex]
For more info, check out polar to cartesian conversion formulas.
