How do I convert this to polar form?

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answers:
Nzewi Ernest Kenechukwu say: Let the vector 6i - 3j be v, i.e.

v = 6i - 3j -- (1)

To convert (1) from its vector form to its polar form (r ∠ ϴ), where r is the magnitude and ϴ is the principal angle, we evaluate thus:

For the magnitude:
|v| = |6i - 3j|, where |v| is r. Then
r = |6i - 3j| -- (2)
Recall from vectors, the formula:
|ai + bj| = √(a² + b²) -- (3)
From (3), we have (2) to be expressed as
r = √[6² + (-3)²]
r = √(36 + 9)
r = √45
r = 3√5 -- (4)

For the principal angle:
Recall in vectors, the formula
ø = arctan (b/a) -- (5) where b is the component vector of the y-axis & a is the component vector of the x-axis. From (5), we evaluate thus
ø = arctan (- 3/6)
ø = - 26.57º -- (6)

Since i (which is the x-axis) is positive and j (which is the y-axis) is negative, then the vector v falls in the 4th Quadrant of a defined unit circle. From (6), the principal angle is evaluated thus:
ϴ = - 26.57º + 360º
ϴ = 333.43º -- (7)

From (5) & (6), we therefore conclude that
6i - 3j = (3√5 ∠ 333.43º) ...Ans.
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Pinkgreen say: Let vector a=6i-3j, then
|a|=sqr[6^2+(-3)^2]
=>
|a|=3sqr(5)
tanA=-3/6=-1/2
=>
A=-26.565* or 333.435*
approximately.
Thus,
a=3sqr(5)[cos(26.565*)i-
sin(26.565*)j]
or
a=3sqr(5)[cos(333.435*)i+
sin(333.435*)j]
or
in polar coordinates:
(3sqr(5), 333.435*).
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Captain Matticus, LandPiratesInc say: 6i - 3j

r^2 = 6^2 + (-3)^2
r^2 = 36 + 9
r^2 = 45
r^2 = 9 * 5
r = 3 * sqrt(5)

tan(t) = -3/6
tan(t) = -1/2
t = arctan(-1/2)
t = -26.565051177077989351572193720453 + 180 * k

k is an integer

<6 , -3> is in Q4, so -26.565 or 333.435
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Ralph say: polar forms sounds cold hahahahahaahahsaa
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Michael say: Magnitude
r² = 6² + (-3)²
r² = 36 + 9
r² = 45
r = √45
r = 3√5
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reference angle
ø = arctan -3/6
ø = -26.57º

Since i is positive and j is negative
this is in Quad IV
ϴ = 360 - 26.57
ϴ = 333.43º
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3√5 ∠333.43º <–––––––
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