Does the Dot Product of Two Non-Zero Vectors Bisect the Angle Between Them?

[ProPatto16]

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, that is, divides it in half.

Homework Equations

none?

The Attempt at a Solution

dont know where to start manipulating?

 

[HallsofIvy]

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, that is, divides it in half.

Homework Equations

none?

Since the question is about the angle between two vectors, I think I would be inclined to use $u\cdot v= |u||v|cos(\theta)$ so that
$$cos(\theta)= \frac{u\cdot v}{|u||v|}$$

to show that the angle between a and c is the same as the angle between b and c.

The Attempt at a Solution

dont know where to start manipulating?

Since c= |a|b+ |b|a, $a\cdot c= |a|a\cdot b+ |b||a|^2$ and $b\cdot c= |a||b|^2+ |b|a\cdot b$.

Further,
$$|c|= \sqrt{(|a|b+ |b|a)\cdot (|a|b+ |b|a)}= \sqrt{2|a|^2|b|^2+ 2|a||b|a\cdot b}$$

 

[ProPatto16]

then the next step is:

[tex]\sqrt{}2|a|²|b|²+2|a||b||a||b|cos(theta)[/tex]

?

now i need to get rid of the sqrt and somehow end up with a half theta?

if i square both sides i get

c² = 2|a|²|b|² + 2|a|²|b|² cos(theta)

heading in the right direction?

 

[ProPatto16]

that should be sqrt of (2|a|²|b|² + 2|a||b||a||b| cos(theta))

 

[ProPatto16]

then |a|²|b|² = (a.bcoz(theta))² ??

can i make that substitution?