Vector Proof

  • #1

Homework Statement


If
c=|a|b+|b|a where a,b, and c are all non zero vectors, show that c bisects the angle between a and b


Homework Equations





The Attempt at a Solution



I'm taking the approach to prove that the angle between b and c= the angle between c and a

I have written the fact that a[itex]\bullet[/itex]b=|a||b|cos [itex]\theta[/itex]
However, I'm not sure how I can work the equation for c to be able to apply the identity.
 

Answers and Replies

  • #2
lanedance
Homework Helper
3,304
2
as you know they are in the same plane and c will be between a and b
[tex]a \bullet b = |a||b|cos\theta [/tex]
[tex]c \bullet b = |c||b|cos\phi [/tex]

and the show [itex]\phi = 2\theta [/itex]
 
  • #3
I'm not quite sure what you are suggesting.

I can draw the relationship

|a||b|cos [itex]\theta[/itex] = |c||b| cos 2[itex]\theta[/itex]


I can eliminate the |b| from both sides, but I don't know where to go from there, since |c| doesn't seem to help when substituting.
 
  • #4
HallsofIvy
Science Advisor
Homework Helper
41,847
966
Then get rid of "c". You are given that c= |a|b+ |b|a so that c.b= (|a|b+ |b|a).b= |a||b|^2+ |b|a.b.
 
  • #5
I can replace the c[itex]\bullet[/itex]b side with what you've suggested but then how am I supposed to include that it is twice the angle. I loose this ability without the trigonometric function.
 
  • #7
lanedance
Homework Helper
3,304
2
try doing what Halls suggested and let us see what you get
 
  • #8
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
15,131
1,726
I can draw the relationship

|a||b|cos [itex]\theta[/itex] = |c||b| cos 2[itex]\theta[/itex]
Is this actually true?
 
  • #9
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
15,131
1,726
I can replace the c[itex]\bullet[/itex]b side with what you've suggested but then how am I supposed to include that it is twice the angle. I [strike]loose[/strike] lose this ability without the trigonometric function.
You're not supposed to assume it's twice the angle. That's what you're trying to prove!

Note that HallsofIvy's suggestion [itex]\vec{c}\cdot\vec{b} = ab^2+b(\vec{a}\cdot\vec{b})[/itex] has the dot product of [itex]\vec{a}[/itex] and [itex]\vec{b}[/itex] in it. That will introduce [itex]\cos\theta[/itex] into the equation.

You can get rid of [itex]c = \sqrt{\vec{c}\cdot\vec{c}}[/itex] from the righthand side by again using the definition of [itex]\vec{c}[/itex].
 

Related Threads on Vector Proof

  • Last Post
Replies
18
Views
1K
  • Last Post
Replies
18
Views
3K
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
7
Views
1K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
1
Views
1K
Top