### Video Transcript

If vector ππ is equal to five, five root three, then the polar form of vector ππ is what. Is it (A) 10, π; (B) 10, π over two; (C) 10, π over three; or (D) 10, π over six.

In this question, we are given vector ππ in rectangular form and are asked to express it in polar form. We begin by recalling that any vector π written in rectangular or component form π₯, π¦ can also be expressed in polar form π, π, where π is the magnitude or length of the vector and π is the angle that the vector makes from the positive π₯-axis. We can convert from one form to the other using the fact that π₯ is equal to π cos π and π¦ is equal to π sin π. We can also represent this graphically by graphing the vector on the coordinate plane.

In this question, vector ππ is equal to five, five root three. We can create a right triangle to help us calculate the values of π and π as shown. Using the Pythagorean theorem, we have π squared is equal to five squared plus five root three squared. The right-hand side simplifies to 25 plus 75. This means that π squared is equal to 100. And square rooting both sides, we have π is equal to the square root of 100. As the magnitude or length of the vector must be positive, we have π is equal to 10.

Recalling the tangent ratio in right triangles, we have tan π is equal to five root three over five. Both the numerator and denominator are divisible by five. So tan π is equal to root three. We recall from our special angles that tan of 60 degrees is equal to root three. And since 180 degrees is equal to π radians, then the tan of π over three radians must also be equal to root three. This gives us our value of π equal to π over three.

An alternative method would be to take the inverse tangent of both sides of our equation. And ensuring weβre in radian mode, the inverse tan of root three gives us π over three radians. We now have the polar form of vector ππ. It is 10, π over three. And from the four given options, this is the same as option (C).

Had we used the fact that π₯ is equal to π cos π and π¦ is equal to π sin π, we would begin by substituting five for π₯ and five root three for π¦. We would then have a pair of simultaneous equations that we could once again solve to calculate the values of π and π. Whilst this is a perfectly valid method, it would be more time-consuming in this case.