1. Apr 6, 2016

### Amine_prince

i have two unitary vectors in space U and V , U and V are perpendicular .
W is a vector that verifies W ^ V = U - W .
the following resolution is incorrect , i wan't to understand why :

we use (o,U,V,(U^V)) . components of U (1,0,0) , V(0,1,0) , W(a,b,c) where a , b and c are real numbers .
components of W ^ V ( c , 0 , a) . and U - W ( 1-a , -b, -c)
where is the error here ?

2. Apr 6, 2016

### stevendaryl

Staff Emeritus
I'm not sure about your notation: ^ means vector cross-product? I've always used $\times$

If so, then you are on the right track. $U, V, U \times V$ can be used as an orthonormal basis. So we can write $W$ as a linear combination:

$W = a U + b V + c (U \times V)$

Then $W \times V = U - W$ becomes:

$(a U + b V + c (U \times V)) \times V = U - a U - b V - c (U \times V)$

Now, we use the rules:
$X \times X = 0$
$(X \times Y) \times Z = (X \cdot Z) V - X (Y \cdot Z)$

where $\cdot$ is the vector scalar product.

Applying these rules gives us:
$a U \times V + 0 + c (U \cdot V) V - c U (V \cdot V) = U - a U - bV - c (U \times V)$

This simplifies to:
$a U \times V - c U = (1-a)U - b V - c(U \times V)$

So if you just pair up the corresponding orthogonal vectors, this gives three equations:
1. $a = -c$
2. $0 = -b$
3. $-c = (1-a)$

3. Apr 6, 2016

### Amine_prince

thank you sir :)

4. Apr 6, 2016

### stevendaryl

Staff Emeritus
Looking at what you wrote, I think that your problem is that

$(a, b, c) \times (0,1,0) = (-c, 0, a)$

5. Apr 6, 2016

### Amine_prince

yes , i missed the sign . and flipped the numbers by mistake .