Vector Reflection

**dimatt** · April 11th 2006, 11:31 AM

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?

**earboth** · April 12th 2006, 09:12 PM

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

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

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

My "solution" looks a little bit too easy.

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

Greetings

EB

**CaptainBlack** · April 13th 2006, 02:17 AM

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

Parrallel component:

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

Othogonal component:

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

Hence:

$ \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:

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

which is the form required.

RonL

**dimatt** · April 14th 2006, 05:35 PM

Wow, thanks guys. I totally forgot about the orthogonal vector

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

**dimatt** · April 17th 2006, 07:06 PM

??

**CaptainBlack** · April 17th 2006, 07:54 PM

Originally Posted by dimatt

Wow, thanks guys. I totally forgot about the orthogonal vector

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 $b$ flipped
from one side of $b$ to the other, that is it is reflecting the orthogonal
component in $b$ .

RonL

**dimatt** · April 18th 2006, 11:32 AM

Oh, i see where i went wrong. I kept looking at c as the projection of a on b; for some reason

Thanks for all of the help.

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