# Vector angle in multiple planes

• Jan 23rd 2007, 07:09 PM
thedoge
Vector angle in multiple planes
How would I go about solving a problem that wanted you to find the angle between two vectors of equal distance given the length of these two vectors and the sum of the two vectors (let's say #j).

You can throw out any number for distance and the sum of the two vectors. I want to see how this is done is all.

edit:
I'll throw in some numbers if that helps. These aren't my actual numbers, but maybe it'd be easier to explain.

Two vectors of equal magnitude 60.
The sum of the two vectors is 20j.
What is the angle between them?
• Jan 23rd 2007, 07:34 PM
ThePerfectHacker
Quote:

Originally Posted by thedoge
How would I go about solving a problem that wanted you to find the angle between two vectors of equal distance given the length of these two vectors and the sum of the two vectors (let's say #j).

You can throw out any number for distance and the sum of the two vectors. I want to see how this is done is all.

For simplicity let us work in 2 dimensions.

You have to vectors (non-zero),
$\bold{u}=$
$\bold{v}=$
You are given the distance $s$.

Thus,
$u_1^2+u_2^2=s^2$ *
$v_1^2+v_2^2=s^2$ **

You are also given their sum,
$+=$
Where $k_1,k_2$ are given.
Thus,
$u_1+v_1=k_1$ (1)
$u_2+v_2=k_2$ (2)

Square (1) and Square (2),
$u_1^2+2u_1v_1+v_1^2=k_1^2$
$u_2^2+2u_2v_2+v_2^2=k_2^2$

$(u_1^2+u_2^2)+2u_1v_1+2u_2v_2+(v_1^2+v_2^2)=k_1^2+ k_2^2$
From * and ** we have,
$2u_1v_1+2u_2v_2+2s^2=k_1^2+k_2^2$
Divide by two,
$\boxed{u_1v_1+u_2v_2=\frac{k_1^2+k_2^2}{2}-s^2}$ (5)

Now, the angle between these vectors are found by,
$\cos \theta = \frac{|\b{u} \cdot \b{v}|}{s}$
Substitute (5) into this equation because that is the dot product.

$\cos \theta = \left|\frac{k_1^2+k_2^2}{2s}-s \right|$
• Jan 23rd 2007, 07:59 PM
thedoge
Thanks. I managed to do it with trig and drawing it out, but your method helped more.