Hey,
where is the problem in the following calculation?
What exactly does your <x, y> notation mean? You’re using it a lot but I don’t understand what it is.
You used it twice in the first line, where you defined cos(theta).
Are you trying to derive the projection of vector ‘x’ onto vector ‘y’?
The notation <x, y> is what Prof Serrano uses in this video to denote the dot product between two vectors.
But the algebra doesn’t work in what you showed above. If you have this:
<x, y> = |x| \cdot |y| \cdot cos(\theta)
Then how do you get to this:
<x, y> = |x| \cdot |y| \cdot \displaystyle \frac {<x, y>}{|x|}
That doesn’t make any sense, right? By applying 8th grade algebra, here’s what I get:
\displaystyle \frac {<x, y>}{|x|} = |y| \cdot cos(\theta)
So how does your result fit with the initial formula? Well, unless you have some reason to know that |y| = 1, but that is not true in general.
Below is the problem:
You can see that it is a problem if you consider this triangle:
Obviously, the adjacent length in my triangle and the adjacent length in your triangle cannot be the same.
Cheers,
Raymond
I guess you took it that way because someone told you that dot product can be interpreted as projection, but I am sure you need to pay closer attention to what they said because not any dot product is projection. You probably want to review that part if it was covered in the lecture, and think about that again.
Another thing that you need think about is, given your triangle,
the adjacent length should not change even if I lengthen vector y by 100 times, however, <x, y> will change if I lengthen vector y. How are you going to resolve this contradiction?
Omg, yes you are right.
So I thought:
<u, v> = |u'|
while in reality, <u, v> = |u'| * |v|, which is precisely what my initial calculation yielded.
1 Like
That’s great!
Cheers!
Raymond