# Another vector problem

1. Feb 22, 2005

### Perion

Hi. I am to show that for any non-zero vectors a, b, and c, if
c = |b|a + |a|b
then vector c bisects the angle theta between vectors a and b. [0 < theta <= Pi]

I used the vectors' component forms and let a = <a1,a2> and b = <b1,b2> and then worked out the equations for the cosine of the angles between (1) a and c, and (2) c and b. I then tried to determine if (1) and (2) were equal. I used the equations

(1) cos(phi1) = (a.c)/|a||c| and
(2) cos(phi2) = (c.b)/|c||b| [a.c and c.b are dot products]

Unfortunately I came up with some rather long, complicated, component expressions which proved quite cumbersome to test for equality. I suspect that James Stewart (my textbook author) had an easier method in mind (maybe using projections) that didn't involve such algebraic gymnastics. Am I missing some obviously simple method to go about this?

Thanks,
Perion

2. Feb 22, 2005

### matt grime

Why do this in 2-d only?

How do you find the angle between two vectors using the dot product? And whats cos(2t) in terms of cos(t)?

If you just use those carefully (no need to do componentsin 2d) it drops out, I imagine.

For example |c|^2=c.c = 4|b|^2|a|^2, so |c|= 2|a||b|

3. Feb 22, 2005

### dextercioby

$$\vec{c}=|\vec{b}|\vec{a}+|\vec{a}|\vec{b}$$ (1)

Take scalar product with itself and obtain:

$$|\vec{c}|^{2}=2|\vec{a}|^{2}|\vec{b}|^{2}[1+\cos(\vec{a}\vec{b})]=4|\vec{a}|^{2}|\vec{b}|^{2}\cos^{2}\frac{(\vec{a}\vec{b})}{2}$$ (2)

Take square root & obtain:

$$|\vec{c}|=2|\vec{a}|\vec{b}|\cos\frac{(\vec{a}\vec{b})}{2}$$ (3)

Consider the scalar product:

$$\vec{a}\cdot\vec{c}=|\vec{a}||\vec{c}|\cos(\vec{a}\vec{c})=|\vec{b}|\vec{a}|^{2}+|\vec{b}|\vec{a}|^{2}\cos(\vec{a}\vec{b})$$ (4)

,where i made specific use of (1).From (4) it follows simply (check the trick with the half-angle formula,as in (2)),that:

$$|\vec{c}|=2|\vec{a}||\vec{b}|\frac{\cos^{2}\frac{(\vec{a}\vec{b})}{2}}{\cos(\vec{a}\vec{c})}$$ (5)

Set equal (4) & (5) and,after the simplifications,u get:

$$\cos(\vec{a}\vec{c})=\cos\frac{(\vec{a}\vec{b})}{2}$$ (6)

Use the fact that the angles are in the interval $(0,\pi)$ and we can make use of the "arccos"...You get simply:

$$(\vec{a}\vec{c})=\frac{(\vec{a}\vec{b})}{2}$$ (7)

,which is just what u were supposed to prove...

Daniel.

P.S.The proof is very general.It makes no reference to any #of dimensions or a base in the vector space...

Last edited: Feb 22, 2005
4. Feb 22, 2005

### mathman

There is a simple approach involving only elementary geometry. |a|b and |b|a are the same length. These form adjacent sides of a rhombus. c is a diagonal starting at the point of origin of these vectors. Therefore it must bisect the angle formed by these sides.

5. Feb 22, 2005

### Perion

Oh wow! Yeah - thanks for the insights. I tried manipulating my expressions in order to use the half angle trig identity but made a dumb mistake so went the idiotic component route. Actually, I shouldn't have assumed just 2D since the problem didn't specify any particular number of dimensions. I just didn't want to deal with more than 2D since the 2D component form resulted in expressions that sucked more enough for my taste.

Matt and dextercioby - your approach is the way that I wanted to go (and suspected the author desired) and now it's also why I'm kicking myself in the butt for abandoning

Thank you very much!!!

Perion

6. Feb 23, 2005

### mathman

The problem is intrinsically 2-D. The two vectors define a plane, which also contains their sum vector.