Video: Finding the Components of a Vector Given in Polar Form

If πŽπ‚ = ⟨4√3, 3πœ‹/4⟩ is the position vector, in polar form, of the point 𝐢 relative to the origin 𝑂, find the π‘₯𝑦-coordinates of the point 𝐢.


Video Transcript

If πŽπ‚ equals four root three, three πœ‹ over four is the position vector in polar form of the point 𝐢 relative to the origin 𝑂, find the π‘₯𝑦-coordinates of the point 𝐢.

Okay, so if this is the π‘₯𝑦-plane, here at the origin is point 𝑂, and then this vector πŽπ‚ tells us the coordinates of a point 𝐢 relative to this origin. We’re told further that this position vector is given to us in polar form. Written this way, the first value is a radial distance, often called π‘Ÿ. And the second value is an angle measured with respect to the positive π‘₯-axis.

In general then, if we had a point located in the π‘₯𝑦-plane and we knew the radial distance from the origin of that point, as well as the angular distance from the positive π‘₯-axis, then we would know the position vector of that point in polar form. We know though that another way to show where this point is located is to give its π‘₯- and 𝑦-coordinates.

When we think about how to calculate these values using π‘Ÿ and πœƒ, this sketch shows us that since the π‘₯-coordinate of our point is equal to this length here and that length makes up a side of a right triangle that is adjacent to πœƒ, that means that π‘₯ is equal to π‘Ÿ times the cos of πœƒ. And likewise, the 𝑦-coordinate of this point, which is equal to this vertical distance here, corresponds to the side of the triangle opposite our angle πœƒ. This means that the 𝑦-coordinate of our point is equal to π‘Ÿ times the sin of πœƒ.

So getting back to our question, to solve for the π‘₯𝑦-coordinates of point 𝐢, we can say that π‘₯ equals π‘Ÿ cos πœƒ and 𝑦 equals π‘Ÿ sin πœƒ, where our given π‘Ÿ-value is four times the square root of three and πœƒ is three πœ‹ over four. We note that this angle three πœ‹ over four bisects the second quadrant. Therefore, the sine of this angle will be positive, while the cosine will be negative. And that makes sense because, in the second quadrant, our π‘₯-value is less than zero, while our 𝑦-value is greater than zero. The cos of three πœ‹ over four equals negative the square root of two over two, while the sine of this angle has the same magnitude but is positive.

When we substitute in these values and multiply through, we find that π‘₯ equals negative two times the square root of six, while 𝑦 is positive two root six. On our graph, these points would look like this, indicating that this point is point 𝐢. The π‘₯𝑦-coordinates of this point are negative two root six, positive two root six.

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