How do I convert this to polar form?

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 yaxis & a is the component vector of the xaxis. From (5), we evaluate thus
ø = arctan ( 3/6)
ø =  26.57º  (6)
Since i (which is the xaxis) is positive and j (which is the yaxis) 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.

Pinkgreen say: Let vector a=6i3j, 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*).

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

Ralph say: polar forms sounds cold hahahahahaahahsaa

Michael say: Magnitude
r² = 6² + (3)²
r² = 36 + 9
r² = 45
r = √45
r = 3√5

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

3√5 ∠333.43º <–––––––

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