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Let's look at going
between representations.

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We'll start in i,
j, k representation

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going to the magnitude
and angle representation.

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So if we have our vector,
the arbitrary vector

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could be written as Ax
i hat plus Ay j hat.

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In this particular case,
we're given a vector

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00:00:19,770 --> 00:00:22,650
minus 2i hat plus 3 hat.

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00:00:22,650 --> 00:00:25,190
And we'd like to go to
the magnitude and angle

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representation.

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So, first of all, let's draw
it out on our grid like this.

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And let's find the magnitude.

12
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So the magnitude we can
find through the Pythagorean

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theorem.

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It's just the square root
of the x-component squared

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plus the y-component squared.

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And now we can find the angle.

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The angle, the
tangent of the angle,

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is just equal to the
y-component divided

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by the x-component like this.

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And so in this way, we
can solve for the angle.

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Now, let's practice
going back the other way.

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If we're given a vector
whose magnitude is 2

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and whose angle from the
x-axis is 30 degrees,

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like this one here,
then the x-component

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is just the magnitude times
the cosine of the angle,

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so 2 times the cosine of 30.

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And the y-component
is just the magnitude

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times the sine of the angle,
so 2 times the sine of 30.