# Vector Reflection

• Apr 11th 2006, 11:31 AM
dimatt
Vector Reflection
First of all, sorry if i posted this in the wrong forum. This is for my Discrete preparation course for univerisity and there was no Discrete forum under the highschool heading :(

Anyways here is the question:

Vectors 'a' and 'b' are drawin tail-to-tail. Vector 'c' is the reflection of 'a' in the line containing vector 'b'. Express 'c' as a linear combination of 'a' and 'b'.

I tried showing it through the use of the projection of a vector formula but I keep getting stuck. Could any of you please offer some guidance or a suggestion on where to start?
• Apr 12th 2006, 09:12 PM
earboth
Quote:

Originally Posted by dimatt
...
Anyways here is the question:
Vectors 'a' and 'b' are drawin tail-to-tail. Vector 'c' is the reflection of 'a' in the line containing vector 'b'. Express 'c' as a linear combination of 'a' and 'b'.

Could any of you please offer some ... suggestion on where to start?

Hello,

I've attached an image to demonstrate, what I've done:

1.: $\displaystyle |\vec{a}|=|\vec{a'}|$

2.: $\displaystyle \vec{a}+\vec{a'}=c \cdot \vec{b}$
(c is a constant factor)

Therefore: $\displaystyle \vec{a'}=c \cdot \vec{b}-\vec{a}$

My "solution" looks a little bit too easy.

The other possibility I can think of is, that you use matrices.

Greetings

EB
• Apr 13th 2006, 02:17 AM
CaptainBlack
Quote:

Originally Posted by dimatt
First of all, sorry if i posted this in the wrong forum. This is for my Discrete preparation course for univerisity and there was no Discrete forum under the highschool heading :(

Anyways here is the question:

Vectors 'a' and 'b' are drawin tail-to-tail. Vector 'c' is the reflection of 'a' in the line containing vector 'b'. Express 'c' as a linear combination of 'a' and 'b'.

I tried showing it through the use of the projection of a vector formula but I keep getting stuck. Could any of you please offer some guidance or a suggestion on where to start?

Resolve $\displaystyle \bf a$ into componets parallel and orthogonal to $\displaystyle \bf b$.

Parrallel component:

$\displaystyle (\bf a . \bf b) \hat{\bf b}$

Othogonal component:

$\displaystyle \bf a - (\bf a . \bf b) \hat{\bf b}$

Hence:

$\displaystyle \bf c=(\bf a . \bf b) \hat{\bf b}-(\bf a - (\bf a . \bf b) \hat{\bf b})=2(\bf a . \bf b) \hat{\bf b}-\bf a$

Which may be rewritten:

$\displaystyle \bf c=2\ \frac{\bf a . \bf b}{|\bf b|}\ \bf b-\bf a$

which is the form required.

RonL
• Apr 14th 2006, 05:35 PM
dimatt
dimatt
Wow, thanks guys. I totally forgot about the orthogonal vector :p

edit- Why do you subtract the parallel from the orothogonal? Wouldnt that just give you the length of 'a'?
• Apr 17th 2006, 07:06 PM
dimatt
??
• Apr 17th 2006, 07:54 PM
CaptainBlack
Quote:

Originally Posted by dimatt
Wow, thanks guys. I totally forgot about the orthogonal vector :p

edit- Why do you subtract the parallel from the orothogonal? Wouldnt that just give you the length of 'a'?

It gives a vector with the orthogonal component about $\displaystyle b$ flipped
from one side of $\displaystyle b$ to the other, that is it is reflecting the orthogonal
component in $\displaystyle b$.

RonL
• Apr 18th 2006, 11:32 AM
dimatt
Oh, i see where i went wrong. I kept looking at c as the projection of a on b; for some reason :o Thanks for all of the help.