Linear Algebra- Vector proof

[sdoyle]

Homework Statement

Prove that if a vector u, is perpendicular to both v and w, then u is also perpendicular to v+w. More generally, show that u (perp) (sv+tw) for all scalars s and t.

Homework Equations

I was thinking that the cross product would be relevant.

The Attempt at a Solution

I'm not quite sure where to start. I tried to use w[-wy, wx] and so forth, but I honestly don't understand their full meanings. Any help would be appreciated

 

[Dick]

I think you ought to be thinking about the dot product. How do you express the notion two vectors are perpendicular in terms of the dot product?

 

[rock.freak667]

If u is perpendicular to v and w, then v and w should lie on the same plane, correct? Thus if v+w, is also perpendicular to u, then what can you say about the vectors, v,u and v+w ?


EDIT: I believe Dick's method is more feasible than what I was trying to do.

 

[sdoyle]

that u^.(v+w)=0?

 

[Dick]

Right. u.(v+w)=0 says u and v+w are perpendicular. Does that follow from u.v=0 and u.w=0?

 

[sdoyle]

Yes? Because both v and w would have to be perpendicular to u? I'm not really sure though...

 

[Dick]

The problem says "u is perpendicular to both v and w". That means u.v=0 and u.w=0. What can you conclude about u.(v+w) and why?

 

[sdoyle]

I really don't know.. I'm sorry. I don't think that I have enough background knowledge, we've only had one lecture.

 

[rock.freak667]

I really don't know.. I'm sorry. I don't think that I have enough background knowledge, we've only had one lecture.

Check the http://en.wikipedia.org/wiki/Dot_product#Properties"

 

[NoMoreExams]

Write out u.v + u.w and u.(v+w)

 

[sdoyle]

u.v=-u.w? Because of distributive laws?

 

[rock.freak667]

u.v=-u.w? Because of distributive laws?

By the distributive laws, can you expand u.(v+w)?

 

[NoMoreExams]

You should be able to see that u.v + u.w = u.(v+w) just by writing out the terms on each side

 

[Dick]

u.v=-u.w? Because of distributive laws?

No. Because of distributive law u.(v+w)=u.v+u.w. Doesn't that look more like a 'distributive law'?

 

[sdoyle]

right...

 

[sdoyle]

but we know that u.(v+w)=0 . Wouldn't that imply that u.v+u.w=0?

 

[rock.freak667]

but we know that u.(v+w)=0 . Wouldn't that imply that u.v+u.w=0?

Actually, you knew that u.(v+w)=u.v + u.w, from the question, you deduced that u.v=0 and u.w=0
So u.(v+w)=0, which means?

 

[sdoyle]

that the vector u is perpendicular to v+w?

 

[rock.freak667]

that the vector u is perpendicular to v+w?

Yes.

Now you want to prove that u is perpendicular to (sv+tw) for all scalars s and t.

Now consider what u.(sv+tw) expands out to be.

 

[Dick]

but we know that u.(v+w)=0 . Wouldn't that imply that u.v+u.w=0?

You don't know u.(v+w)=0. That's what you are trying to prove. What you know is that u.v=0 and u.w=0.

 

[sdoyle]

right but u.(v+w)=u.v+u.w
=0+0
= 0
In terms of the scalars any number multiplied by zero will yield zero