B Error in Vector Addition: U & V Perpendicular

[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 ?

 

[stevendaryl]

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 ?

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)

 

[Amine_prince]

thank you sir :)

 

[stevendaryl]

thank you sir :)

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

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

 

[Amine_prince]

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