Find unknown vector X if these relations hold true

[Purcolator]

Homework Statement

If an unknown vector X satisfies the relation
X · b = β
X × b = c
express X in terms of β, b, and c.

Homework Equations

X · b = |X||b|cos(θ)
X × b = |X||b|sin(θ)

The Attempt at a Solution

I don't know where to start... :( someone pls give me a hint

 

[Stephen Tashi]

Homework Equations

X · b = |X||b|cos(θ)
X × b = |X||b|sin(θ)

For the cross product, you need notation to say "the magnitude of " | X \times b | = ...

One approach to this problem would be to represent the vectors in their 3 components (e.g. (X_x,X_y,X_z)) but I think that would be a big mess.

Another approach is to assume that "express X in terms of \beta,b,c can be fulfilled by a treating those quantities as single symbols. With that approach we can write X = P + Q with P, Q vectors as long as they are expressed in terms of the symbols \beta, b, c. Begin by letting P be the component of X that in in the same direction as vector b. The direction of b is expressed as the unit vector \frac{b}{|b|}. Can you find the component of X in that direction? Perhaps you have studied how to "project" a vector onto another vector.

 

[Purcolator]

OK that makes sense! so I'm guessing Q would be the projection of X onto c? However I'm stuck from here because i cannot express that projection in terms of correct variables (don't know A dot C)... am i wrong?

 

[Stephen Tashi]

OK that makes sense! so I'm guessing Q would be the projection of X onto c?

It wouldn't be the projection on c because c is perpendicular to the plane of X and b. To get something in the same plane as X and b, try the vector (b x c).

 

[Purcolator]

isn't (b x c) = X?

 

[Stephen Tashi]

isn't (b x c) = X?

Suppose b = the unit x vector and X = (1,1,0). Then X x b = c has the direction of (-1) times the unit z vector. And b x c has the direction of the unit y vector.

 

[Purcolator]

OK that makes sense... but how would i express X dot (B x C) when i am formulating the projection? I'm trying to take X out... :/

 

[Stephen Tashi]

OK that makes sense... but how would i express X dot (B x C) when i am formulating the projection? I'm trying to take X out... :/

My guess is that you'll have to use one of the formulas that expresses X dot (b x c) in a different way. ( perhaps the signed volume of a parallelopiped formula in http://en.wikipedia.org/wiki/Cross_product ) See if there is an expression that uses X x b or b x X since you can replace those expressions with c or -c. Don't forget that projecting X on a vector V involves dividing (X dot V) by |V|.

 

[Purcolator]

Thanks so much. i finally got the answer... I appreciate you tremendously!