Vector angle in multiple planes

**thedoge** · January 23rd 2007, 07:09 PM

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?

**ThePerfectHacker** · January 23rd 2007, 07:34 PM

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),
$\bold{u}=<u_1,u_2>$
$\bold{v}=<v_1,v_2>$
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,
$<u_1,u_2>+<v_1,v_2>=<k_1,k_2>$
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),
$u_1^2+2u_1v_1+v_1^2=k_1^2$
$u_2^2+2u_2v_2+v_2^2=k_2^2$

Add these two equations,
$(u_1^2+u_2^2)+2u_1v_1+2u_2v_2+(v_1^2+v_2^2)=k_1^2+ k_2^2$
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.

$\cos \theta = \left|\frac{k_1^2+k_2^2}{2s}-s \right|$

**thedoge** · January 23rd 2007, 07:59 PM

Thanks. I managed to do it with trig and drawing it out, but your method helped more.

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