Projection Vectors

In that same worksheet that the question about closest approach came from, there was also a question about projection vectors.

Find the projection of 4i - 3j onto the vector 12i + 8j.

First I had to find out what a projection vector was. So I asked my Spec teacher.

With the vectors 4i - 3j and 12i + 8j, as illustrated below:

The projection vector is the purple part on the next diagram:

The projection of a vector onto another is the effect of that vector in the direction of the other vector. (I'm not 100% sure what this means, but I'll get there eventually.)

As can be seen from the diagram, you can draw a line between the 4i - 3j vector and a particular point on the 12i + 8j vector which is perpendicular to the 12i + 8j vector. (It's a bit like those closest approach questions.) The particular point where the lines intersect has a position vector identical to the projection vector of 4i - 3j on 12i + 8j. Therefore, finding the projection vector is all about finding this particular point.

Hopefully that wasn't too confusing.

As you can see above, we have a nice little right-angled triangle. Here it is, rotated because it's easier to understand (in my opinion, anyway).

Let us call 4i - 3j b, and the purple line x.

Using simple trigonometry cos(theta) = |x|/|b|. Therefore |x| = |b| cos(theta).

Now we need to work out cos(theta). Remember, theta in this case is the angle between 12i + 8j (which I'm going to call "a," by the way), and 4i - 3j (which I called "b" not too long ago).

Let's use the scalar product equation to find cos(theta). a(dot)b = |a||b|cos(theta) can be rearranged to cos(theta) = (a(dot)b)/(|a||b|).

Therefore x = (|b|(a(dot)b))/(|a||b|). The |b| then cancels out to give (a(dot)b)/|a|.

This only gives us the magnitude of line x, however. We also need to find the direction, which is in the same direction as line a.

As I'm sure you probably already know if you're also doing 3AB Spec, if you want to find a vector of a certain magnitude in the same direction of another vector, you just need to multiply the desired magnitude by the unit vector (since the unit vector has a magnitude of 1). Therefore, we need to multiply the magnitude of x by the unit vector of a.

The unit vector of a can be represented by the symbol â, or by a/|a| (i.e. vector a divided by the magnitude of a).

Therefore the projection vector is as follows:

((a(dot)b)/|a|)/(a/|a|)

which can be simplified to:

which can be simplified even further to:

(â.b)â

Hopefully my explanations weren't too complex and you now at least have some kind of understanding of projection vectors! Yeah... it was a real Eureka moment yesterday when I understood...