Hard vector proof using dot product

[hollicrombie&stitch]

Hello! I'm looking for help to solve part b of this problem (the answer to part a is " a - 4b ").
I know that because it is a square, the diagonals are equal in magnitude and perpendicular, meaning the dot product would be equal to zero, but was unable to obtain a version with just |a| and |b| and not |ab|, or something that would otherwise allow me to derive |a| = 2|b|²
This is my first post so sorry if I put it under the wrong category. Please give a senior high-school level answer otherwise my brain might explode. Thanks!

 

[Deleted member 4993]

Hello! I'm looking for help to solve part b of this problem (the answer to part a is " a - 4b ").
I know that because it is a square, the diagonals are equal in magnitude and perpendicular, meaning the dot product would be equal to zero, but was unable to obtain a version with just |a| and |b| and not |ab|, or something that would otherwise allow me to derive |a| = 2|b|²
This is my first post so sorry if I put it under the wrong category. Please give a senior high-school level answer otherwise my brain might explode. Thanks!
View attachment 17297

Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING

Please share your work/thoughts about this assignment.

 

[pka]

Hello! I'm looking for help to solve part b of this problem (the answer to part a is " a - 4b ").
I know that because it is a square, the diagonals are equal in magnitude and perpendicular, meaning the dot product would be equal to zero, but was unable to obtain a version with just |a| and |b| and not |ab|, or something that would otherwise allow me to derive |a| = 2|b|²
This is my first post so sorry if I put it under the wrong category. Please give a senior high-school level answer otherwise my brain might explode. Thanks!
View attachment 17297

I agree with the answer to part a. I wonder if part b is correct?
Do you know that \(\vec{{\bf a}}\cdot\vec{{\bf a}}=\|\vec{{\bf a}}\|^2~?\).
You are correct \((3\vec{{\bf a}}-2\vec{{\bf b}})\cdot(\vec{{\bf a}}-4\vec{{\bf b}})=0\)
AND \(\|3\vec{{\bf a}}-2\vec{{\bf b}}\|=\|\vec{{\bf a}}-4\vec{{\bf b}}\|\)
See what you can do with those.

 

[hollicrombie&stitch]

Here's all the working I got through, I tried several different dot product expansions and kept getting stuck when I was unable to cancel out |ab| or a⋅b to just have something in terms of |a| and |b|. I was able to determine all four sides of the square and both diagonals in terms of a and b. Even tried the trusty quadratic formula! Sorry for not posting this earlier. Thanks again for the help.

 

[pka]

Here's all the working I got through, I tried several different dot product expansions and kept getting stuck when I was unable to cancel out |ab| or a⋅b to just have something in terms of |a| and |b|. I was able to determine all four sides of the square and both diagonals in terms of a and b. Even tried the trusty quadratic formula! Sorry for not posting this earlier.

I have looked over your work. I realize that our notation is a bit different. \(\vec{{\bf a}}\) is the vector a and \(\|\vec{{\bf a}}\|\) is its norm(length).
If we take \(OPQR\) is a square then as you have shown
\((2\vec{{\bf a}}-3\vec{{\bf b}})\cdot(\vec{{\bf a}}+\vec{{\bf b}})=0\\2\vec{{\bf a}}\cdot\vec{{\bf a}}-\vec{{\bf a}}\cdot\vec{{\bf b}}-3\vec{{\bf b}}\cdot\vec{{\bf b}}=0\\2\|\vec{{\bf a}}\|^2-3\|\vec{{\bf b}}\|^2=\vec{{\bf a}}\cdot\vec{{\bf b}}\)
I really wonder if part b is a typo?

 

[hollicrombie&stitch]

Yeah we had a group of us trying to solve it for a whole hour to no avail. Might have to take it to my teacher next lesson. Thanks for taking the time to look at this!

 

[pka]

Yeah we had a group of us trying to solve it for a whole hour to no avail. Might have to take it to my teacher next lesson. Thanks for taking the time to look at this!

I hope that your group will continue to work on this. It is entirely possible that I do not follow the notation. But I do follow the work you posted. I tried to reverse engineer question b. No luck. So when you ask the instructor, please point that out. Also please post the response here.

 

[hollicrombie&stitch]

I don't have class for a couple of days, but when I get the answers I'll definitely post them here

 

[hollicrombie&stitch]

The solution is to expand (2a - 3b) ⋅ (a + b) and (3a - 2b) ⋅ (a - 4b), get an expression of a⋅b in terms of |a| and |b| for one of the equations, then sub into the other and show that |a| = 2|b|^2
Thanks again for all the help!