And now, the follow up - vectors.
So, before we very explicitly only talked about vectors in terms of dynamics. Today, we're going to talk about them in terms of themselves, and just go over everything you can do in the vectors topics.
So, let's say you have your vector, a = xi + yj + zk, or b=ui+vj+wk. It has two properties - direction and magnitude. Direction can be determined just by drawing your vector on a graph. Another way you might choose to do this is in terms of the angle that the vector makes with each axis - more on this later. If you wanted to define the magnitude, however, you have to do this in terms of pythagoras:
Next, we can add vectors together. If you do this algebraically, it's as easy as adding is with normal numbers. But don't forget, there's also a geometric idea of adding vectors together, which looks a bit like this:
Importantly, this shows that a+b = b+a. Now, people like to assume that stuff like this is always true - it's not, and I'd show you a good example, but I don't want to be the reason for a freak-out just before an exam.
Now, we can also do multiplication and division with vectors - to an extent. If you do this with a vector and a scalar, it's very straight-forward - just pretend you're expanding brackets. And again, don't forget the geometric argument:
This also shows how you'd do subtraction geometrically - instead of thinking of your vector as a - b, think of it as a + (-b) - so you're adding a with negative b.
That's all well and good - but how about multiplying vectors together? This is much harder, and there are (believe it or not) two different ways of multiplying vectors together. In specialist, we only care about the simple form of one of these, and it's called the dot product. It can be evaluated in one of two ways:
Now, remember we talked about finding angles to axes? We can do that with this dot-product. In the case of the x-axis, we make b point in just the direction of the x-axis (so b=ui for this demonstration), and we calculate the dot product in both ways:
Where theta_x is the angle made by a with the x-axis. Now, since they're both equal to the same thing, just calculated in different ways, we can make them equal and solve for our angle:
Now, this bit might be a little confusing - but, it should be obvious that if u is positive, then it can be cancelled out from both sides with no issue. Now, if we want our angle to be with the POSITIVE x-axis, then u is positive, so that's not an issue, and we can just cancel away. This is just something to keep in mind if you copy this - if you use this formula, but you were asked to find the angle made by the NEGATIVE x-axis, your answer is actually going to be 180-the answer we get from this formula. This gives us:
And now you just need to take the inverse cosine! Feel free to this out for the other axis, you might notice a pattern forming.
Now, there's only two things left to go over - the first, is resolving into components. We've actually already done this before - in the dynamics question, instead of making the j point up and i point right, we made them point perpendicular and parallel to the plane. Us multiplying the weight by the sine and cosine of that angle? THAT'S resolving the weight vector into two different components. Now, let's say instead of knowing the angle of the plane, let's say we instead knew that it inclined along the vector b=i+j. Well, we could calculate the angle and then do what we did before, but that would get very complicated if instead the plane was a three-dimensional object. So, let's find a way to do it in terms of b, instead.
Well, we can find out that the angle made with the x-axis is 45 degrees. We also know that the dot-product has a cos in it, and in the dynamics problem, the cos and sin vectors point as shown in the attached figure (note: this is after applying the z-rule. Hopefully this helps with the last lesson, as well, if you didn't understand that part in where the angle ended up!). Using this, we know that we want our final vector to look like acos(theta) (where theta is the angle between the b-vector and the ground) to get the component that points in the j direction in our dynamics diagram. We can actually get this using dot-products:
And this gives us our desired vector length! Now, very confusingly, this is often called the parallel component - and that's because of this image:
See, while this gave us the component perpendicular to our plane, there's actually a second vector that makes the same angle to our weight vector as the plane does - but it points perpendicular to it, and what we've actually done is resolved our weight vector to this second vector. The fact is that the vector maths was developed before the inclined plane problem was, hence why the two don't make sense together - the vector definition is based on what it's always been, but there's no point defining an inclined plane based on this second vector.
However, this does make the next step make sense. Let's say that we re-define b so that it is now i - j - now, this vector is perpendicular to our original b, but points in the direction as this second vector that I mentioned before. This means that now that we have our resolved magnitude, we can multiply this by our new b-vector to get the final vector that is perpendicular to the plane. However, we don't want to include the magnitude of this b-vector, so we also need to divide by its magnitude. This gives us the final formula:
Note that the little hat indicate that we use the unit vector instead of the actual vector, calculated using the handy little equation:
Okay, so, how do we then find the perpendicular component (the one parallel to the plane - to be, you know, confusing and all that)? Well, we could do it in terms of sine, but we don't have a vector formula with a sin(theta) in it. However, we do know from the geometric definition of vector addition that:
parallel projection(a.b/|b|^2 * b) + perpendicular projection (p)=a
We can now solve this to get the perpendicular projection:
Now, I know figuring out which is parallel and which is perpendicular is going to be very confusing - read the question. Normally, VCAA are nice, because they KNOW that the wording is confusing. It's for this reason that inclined plane problems are usually approached by saying "the components parallel to the plane" and "the components perpendicular to the plane", and when doing vector resolutes, they will specifically say "find a in the direction of b" - in this case, we don't need to worry about this "second vector", we can just blindly use the equations we found earlier.
There's also only one more thing left to learn in terms of vectors! That is, independency. We say that a vector a is INDEPENDENT from two vectors b and c if you cannot construct a from scalar multiples of b and c. That is:
If you think of it like this, that means that if the vectors a, b, and c are DEPENDENT, then when you add up some scalar multiple of all three, then you return to where you started - that is, you get the zero vector. So, you can also use the equation:
where k, m, and n are not all 0 at the same time.
Proving things are independent from each other is hard. Instead, what you should do, is try to show that they're dependent - using one of the following two equations:
If you find that there is no solution to the first equation, or instead you show that the only solution to the second equation is k=m=n=0, then that shows that these vectors are independent.