sorry, i have a question that is about vectors, i'm still not quote sure what exactly are vector resolutes, and whats it purpose ( is it purely for finding the shortest distance- is the distance from a point to a line always the shortest if its perpendicular).
Thanks.
I'M BACK AND READY TO RUMBLE.
Okay, so hopefully you're familiar with this picture:
If not, you are now!
So, in the picture, we're projecting the red vector (
u) onto the blue vector (
v). However, before we go into this some more, let's explore the idea of vector components:
If you have a vector
a=3i + j, you know that it moves 3 units in the positive x direction and one unit in the positive y direction. We can then break this vector up into two components - one for the x direction (3i) and one for the y direction (j).
In fact, we can do this for any vector, and the amount of components we can break it into depends on the dimension of the vector space. So,
a is a 2-dimensional vector and can be broken into 2 components. For
b=i-j+2k, we have three components. Of course, we could've said that
a is a three dimensional vector also, and simply doesn't move up or down - in which case, its third component would be 0k.
Now, what if we wanted to change our axis? For example, instead of having the axis measured by vectors i and j, what if they were measured by
and
? Well, all of a sudden, our previous method of finding our two components doesn't work. We note that finding these components is also referred to as "resolving" the vector into "rectangular" components.
So, now we consider the picture above. Let's say that our vector v is the same as our vector x above, and
. Now, to find the projected vector, we use triangular methods. Looking above, we can see that the green vector has length
)
, where theta is the angle between the vectors. Okay, so now we have the length - but a vector also needs a direction, so we need to give it a direction. Of course, the direction of our "axes" is just the unit vector in its direction, which is
(you can calculate the magnitude yourself to confirm.
)
So, this means that the projected vector is
\right)\frac{v}{|v|}=\left(|u|\cos(\theta)\right)\mathbf{\hat{v}})
. However, this means we HAVE to find the angle, which is tedious and annoying - but, we know that
=u.v)
, so we multilpy and divide by the magnitude of v like so:
\right)\mathbf{\hat{v}}=\left(\frac{1}{|v|}u.v\right)\hat{\mathbf{v}}=\left(u.\hat{v}\right)\mathbf{\hat{v}})
Now that you hopefully understand the geometric interpretation, we can discuss the applications. One application is of course finding the shortest distance, which is when the distance between the two vectors/point and one vector is perpendicular. Another is, as I alluded to, the rotation of axis, which has uses if you have an equation such as
(which is actually an ellipse, and can be find by rotation of axis). Resolutes also has applications in physics, such as in mechanics which you will see when you get there. (note: your book/teacher may not make the link when you're resolving force vectors into rectangular components. However, they are very much the same thing)
Aside from the applications in mechanics, VCAA don't seem to have ever assessed applications of resolutes, and applications aren't directly mentioned in the study design.
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