Does |a|*b + |b|*a halve the angle between two random vectors a and b?

[Nikitin]

Hi! please help:
a = vector a
b = vector b

Prove that |a|*b + |b|*a halves the angle between two random vectors a and b.

OK guys I need help proving this. I only go to what is equivalent to high-school in the US ( I'm 17 years old) so no advanced math/computer math please.

Here is what I got: From " |a|*b + |b|*a " we see that |a|*b and |b|*a will always be equally long, but they must always go opposite vertical direction.

BUT, what I must prove for my proof to be legit is this: |(|a|*b)| = |(|b|*a)| so basically that these two vectors will always be equally long.

 

[tiny-tim]

Welcome to PF!

Hi Nikitin! Welcome to PF!

Hint: use a dot product. ​

 

[Nikitin]

hm I think I am about to solve it by simplifying " |(|a|*b)| " as much as possible (and then I will compare it to " |(|b|*a)| " which should be the same as " |(|a|*b)| ")

 

[Nikitin]

tiny-tim, thanks for the welcome! and yeh I used vector-multiplication rules 2 prove that

|(|a|*b)| = |(|b|*a)|

by doing allot of simplifying. In both I got the same results and thus the the theory was confirmed (because (|b|*a) + (|a|*b) will create a new vector, which would obviously then cut the angle between the two old vectors "a and b" exactly in half, if all the vectors start from the same coordinate).

 

[tiny-tim]

Hi Nikitin!

tiny-tim, thanks for the welcome! and yeh I used vector-multiplication rules 2 prove that

|(|a|*b)| = |(|b|*a)|

by doing allot of simplifying.

I'm not sure what your rule 2 is, but you can get that simply from the equation, for any scalar p, |pb| = p|b|

In both I got the same results and thus the the theory was confirmed (because (|b|*a) + (|a|*b) will create a new vector, which would obviously then cut the angle between the two old vectors "a and b" exactly in half, if all the vectors start from the same coordinate).

Sorry, I don't follow that argument …

what's "obvious" about it? ​

 

[Nikitin]

sorry, my English is not very good... I will try to write better and more clear.

2 is an abbreviation for "to".

I haven't learned that rule yet (p, |pb| = p|b|), but now at least I've proven it (hehe).. This problem was supposed to be the hardest I could find about "scalar-products" in my text-book (I am preparing for the exam coming up), so they probably wanted me to think more logical.

(I) Anyway, here are the definitions I used to prove this theory (|a|*b + |b|*a halves the angle between two random vectors a and b):

|a|= Sqrt (a_x² + a_y²)

a = [a_x,a_y]

|b|= Sqrt (b_x² + b_y)²)

b = [b_x,b_y]

Using this, I ended up with

Sqrt (a_x² + a_y²)*(b_x² + b_y²) = | |a| * b |

Sqrt (b_x² + b_y²)*(a_x² + a_y²) = | |b| * a |

(II) Well, if we add two vectors a and b that have exactly same length (|a|=|b|) but opposite vertical directions we will get a new vector, a+b which would cut the angle between the two original vectors in half, if all of the vectors (a, b, a+b) start from same coordinates.

Now, Since |a|*b and |b|*a have identical angle alfa in-between them to vector a and b, if we cut the angle alfa in half for vectors |a|*b and |b|*a with vector |a|*b + |b|*a, then vector |a|*b + |b|*a must do the same to angle alfa between vectors a and b. OK, I know this is very hard to understand, despite my efforts of sounding clear.. but sorry, I am used to a different language when talking about mathematics..

 

[tiny-tim]

Hi Nikitin!

2 is an abbreviation for "to".

Not on this forum …

please see the forum rules on use of English.​

|a|= Sqrt ((a_x)² + (a_y)²) …

oooh no, I'm not reading all those coordinate (x and y) equations …

you should be able to prove this without coordinates …

use the formula |p.q| = |p||q|cosθ

 

[Nikitin]

allrite, thanks for the help mate! I will look it up (I believe we went into this subject in mini-chapter of textbook; "scalar-products and parallel-components").

PS. I edited my post to make it look more readable, as I completely screwed up with the use of ( and )

 

[Nikitin]

Oh, I just read that chapter and we HAVE had about it, but the writers haven't declared it as a rule in my book (probably more about it in next years math, as we don't need that rule for the calculations in this year's physics)..

anyway thanks for the tip and good night, I need to sleep

 

[Rasalhague]

for any scalar p, |pb| = p|b|

I guess you meant

"for any nonnegative scalar, p, |pb| = p|b|"

or

"for any scalar, p, |pb| = |p| |b|.

 

[tiny-tim]

oooh, yes!

thanks, Rasalhague! ​