• Support PF! Buy your school textbooks, materials and every day products Here!

Projection Problem

  • #1

Homework Statement



The projection of the vector V onto (a,b) = (a,b)
The projection of the vector V onto (-b,a) = (-b,a)
Describe V in terms of a and b

Homework Equations





The Attempt at a Solution



I let V=(x,y) then place that into the projection equation for each to get:

[(x,y)(a,b)/(a,b)(a,b)](a,b) = [ax+by/a^2+b^2](a,b)=(a,b)

[(x,y)(-b,a)/(-b,a)(-b,a)](-b,a) = [-bx+ay/b^2+a^2](-b,a)=(-b,a)

After seeing this I realized that:

k(V)=V so k has to equal 1

so both [ax+by/a^2+b^2] and [-bx+ay/b^2+a^2] should eqaul 1

Then I let:
[ax+by/a^2+b^2]=[-bx+ay/b^2+a^2]
the denominators cancel each other out and I'm left with

ax+by=-bx+ay


After that I solved for y to get y=(a^2+b^2-ax)/b
Subsituting that in I get:
[-bx+a[(a^2+b^2-ax)/b]/b^2+a^2]

Reducing and all that other good stuff I finally get x=[a^2+b^2-(a^3/b)+[(ab^2)/2]]/[(a^2)/b]-b]

I am unaware if I am doing this correctly, maybe it's just how it looks that is throwing me off. Any advice or help would be greatly apperciated. THanks
 
Last edited:

Answers and Replies

  • #2
jambaugh
Science Advisor
Insights Author
Gold Member
2,190
238
You had two equations and two unknowns.

This = 1 and That = 1.
You only used one equation
This = That.
Now use either of the original.... they're linear so you could solve them via linear methods but its just as easy in this case to solve one eqn for y in terms of x and substitute it into the other. When you're done verify that both original equations are true.

There is however a quicker way to do this problem... by noting that the two given vectors are orthogonal to each other!!!
 

Related Threads for: Projection Problem

  • Last Post
Replies
3
Views
793
Replies
1
Views
1K
Replies
3
Views
1K
Replies
4
Views
3K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
6
Views
5K
Replies
1
Views
7K
Replies
5
Views
4K
Top