Question Video: Converting a Vector from Rectangular Form to Polar Form Mathematics • First Year of Secondary School

Video Transcript

If 𝐀 equals negative 𝐢 minus 𝐣, then the polar form of 𝐀 is blank. (A) Root two, 𝜋 over four, (B) root two, three 𝜋 over four, (C) root two, five 𝜋 over four, (D) root two, seven 𝜋 over four.

All right, so given this vector 𝐀 in rectangular form, we want to solve for its polar form. Another way we can write 𝐀 in rectangular form is to express it in terms of its 𝑥- and 𝑦-components like this. And now let’s recall that for a vector expressed in polar form, we give it not by its 𝑥-, 𝑦-components, but rather by 𝑟 and 𝜃. Here, 𝑟 is equal to the square root of 𝑥 squared plus 𝑦 squared, and the tan of 𝜃 is equal to 𝑦 divided by 𝑥. Using these relationships, we then have the ability to convert 𝐀 from its rectangular form as given to its polar form.

That polar form, as we’ve seen, is defined by a radial distance and an angle. And 𝑟, we know, is equal to the square root of the 𝑥-component squared plus the 𝑦-component squared. This is equal to the square root of one plus one or the square root of two, so we’ll substitute this result in for 𝑟 in our polar form of 𝐀. But as we look again at our answer options, notice that this doesn’t narrow down the list. All four choices had the same 𝑟-value of the square root of two. So let’s move on to calculating the angle 𝜃 of our vector. As we do this, it will be helpful to sketch our vector on a coordinate plane.

Let’s say that each of these tick marks represents one unit of distance. And since our vector has rectangular components of negative one and negative one, if the tail of the vector was at the origin, then the vector would look like this. And the angle 𝜃, defining its direction, would be measured from the positive 𝑥-axis to the vector. We see then that 𝜃 will be greater than 𝜋 radians but less than three-halves 𝜋. We can now use this relationship here to solve for it. We know that the components of our vector 𝑥 and 𝑦 are both negative one. Here we’ve taken the inverse tan of both sides of our tan of 𝜃 equation, meaning that 𝜃 equals the arc or inverse tan of negative one divided by negative one.

Here’s what’s interesting, though. If we evaluate this inverse tangent on our calculator, the result we get is exactly 𝜋 divided by four. But looking at our sketch of our vector, we know that that can’t be the correct angle for 𝜃. Here, we need to recall the rule that when we calculate an inverse tangent with a negative 𝑥-value in the fraction, then to correctly solve for the angle 𝜃 measured relative to the positive 𝑥-axis, we need to add 𝜋 radians to our result. It’s by doing this that we avoid a possible error in calculating 𝜃. This potential error can be traced back to the tangent function.

But suffice to say whenever we calculate the inverse tangent of a fraction with a negative 𝑥-value, that is, whenever our vector’s in the second or third quadrants, we’ll need to add 𝜋 radians or 180 degrees, as the case may be, to properly solve for the angle relative to the positive 𝑥-axis. 𝜋 over four plus 𝜋 is equal to five over four times 𝜋. And substituting this value in for 𝜃 in our polar form of 𝐀, we see that in polar form 𝐀 is equal to the square root of two, five 𝜋 over four. We find that listed as option (C) among our answer choices. So, to complete our sentence, if 𝐀 equals negative 𝐢 minus 𝐣, then the polar form of 𝐀 is square root of two, five 𝜋 over four.