### Video Transcript

Complete the following. If the vector form of vector ππ
is two π’ plus two π£, then the polar form of vector ππ is what.

We should first note that the
vector weβre looking to find in polar form is not the vector weβve been given in
vector form. Weβve been given vector ππ, and
weβre looking for vector ππ. Before we can find the polar form
of vector ππ, weβll first need to write it down in vector form. Vector ππ can be seen as the
reverse of vector ππ. It will have the same magnitude,
but the opposite direction. So, given that vector ππ is two
π’ plus two π£, then vector ππ will be negative two π’ plus negative two π£, which
we can write more simply as negative two π’ minus two π£.

Now, we need to express this in
polar form. The vector form expressed ππ in
terms of the displacements in the π₯- and π¦-directions, but the polar form
expresses it in terms of the vectorβs magnitude π and its angle π. Letβs sketch a diagram to represent
vector ππ. The vector negative two π’ minus
two π£ represents a movement of negative two in the π₯-direction and negative two in
the π¦-direction. If we position the vector to start
at the origin, we see that it lies in the third quadrant. The length of this vector is given
by π, and the angle π is measured in the counterclockwise direction from the
positive π₯-axis.

The value of π can be calculated
using the Pythagorean theorem as the square root of π₯ squared plus π¦ squared. And the tangent of angle π can be
represented as π¦ divided by π₯. For vector ππ, we can see that
the change in π₯-value is negative two, and the change in the π¦-value is also
negative two. To calculate the magnitude of the
vector, we need to calculate the square root of negative two squared plus negative
two squared. Thatβs the square root of four plus
four, which simplifies to two root two.

Now, tan π is equal to the
π¦-value over the π₯-value, which is negative two over negative two, which
simplifies to one. This means that we need to find the
inverse tan of one to calculate the value of π. Our calculator gives the result
that π is equal to 45 degrees. However, if we look back at our
sketch, we know this isnβt correct; π is much greater than 45 degrees.

The calculator actually gave us the
measure of the acute angle between our vector and the π₯-axis, marked here in
green. It happens to have the same value
as its tangent. So, we need to add this extra 180
degrees to find the true value of π as the measure of the counterclockwise angle
between the positive π₯-axis and our vector ππ. And that gives us 225 degrees.

Itβs always a good idea to sketch
the vector starting from the origin on a pair of coordinate axes, as we did, before
finding the angle for the polar form, just to make sure we correctly take account of
which quadrant it lies in. So, the answer to the question is
that, in polar form, the vector ππ has magnitude two root two and an angle of 225
degrees.