- #1
EDavis
- 2
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Hi there, I had an assignment due today for my advanced maths class, I've already handed it in, but seeing what other people did I began to worry/wonder about what I did.
The question was pretty simple:
For non-zero vectors show that:
|a - b| = |a + b| if and only if venctors a and b are perpendicular.
I defined them both as different single letters, just to make things easier.
a + b = c
a - b = dThen I used the dot product on them both:
c^2 = (a + b) o (a + b)
c^2 = |a + b| |a + b| cos A
c^2 = a^2 + b^2 + 2ab cos A
In this case A is the angle opposite vector c, ie. the angle between vectors a and b.
d^2 = (a - b) o (a - b)
d^2 = |a - b| |a - b| cos B
d^2 = a^2 + b^2 - 2ab cos B
In this case B is the angle opposite vector d, ie. the angle between vectors a and -b.
I then said that in order for |c| to equal |d| the following would have to be true:
cos A = cos B = 0 because otherwise you would have to add/subtract.
Of course it follows from this that A = B = 90 hence a and b are perpendicular if |c| = |d|
I think it's pretty easy to see from that that if they're perpendicular then |c| = |d|, but my maths teachers don't take kindly to the "no duh" argument, they claim that's what proof is all about, so i proved it backwards to show the if and only if statement.
My question is: other people did it using the dot product in a different way to me, does my method make sense? Also, should I have proved it backwards? I could lose marks for having irrelevant lines, but I could've lot lines for missing lines, it's a fine line.
Worried about this because when I did it it was a really easy assignment and I don't want to mess it up :(
Thanks,
Evan
Alright, I removed the tex formatting, as it was too confusing for me.
The question was pretty simple:
For non-zero vectors show that:
|a - b| = |a + b| if and only if venctors a and b are perpendicular.
I defined them both as different single letters, just to make things easier.
a + b = c
a - b = dThen I used the dot product on them both:
c^2 = (a + b) o (a + b)
c^2 = |a + b| |a + b| cos A
c^2 = a^2 + b^2 + 2ab cos A
In this case A is the angle opposite vector c, ie. the angle between vectors a and b.
d^2 = (a - b) o (a - b)
d^2 = |a - b| |a - b| cos B
d^2 = a^2 + b^2 - 2ab cos B
In this case B is the angle opposite vector d, ie. the angle between vectors a and -b.
I then said that in order for |c| to equal |d| the following would have to be true:
cos A = cos B = 0 because otherwise you would have to add/subtract.
Of course it follows from this that A = B = 90 hence a and b are perpendicular if |c| = |d|
I think it's pretty easy to see from that that if they're perpendicular then |c| = |d|, but my maths teachers don't take kindly to the "no duh" argument, they claim that's what proof is all about, so i proved it backwards to show the if and only if statement.
My question is: other people did it using the dot product in a different way to me, does my method make sense? Also, should I have proved it backwards? I could lose marks for having irrelevant lines, but I could've lot lines for missing lines, it's a fine line.
Worried about this because when I did it it was a really easy assignment and I don't want to mess it up :(
Thanks,
Evan
Alright, I removed the tex formatting, as it was too confusing for me.
Last edited:
