# Find unknown vector X if these relations hold true

1. Jan 29, 2015

### Purcolator

• Member warned about posting a problem with no effort shown
1. The problem statement, all variables and given/known data
If an unknown vector X satisfies the relation
X · b = β
X × b = c
express X in terms of β, b, and c.

2. Relevant equations
X · b = |X||b|cos(θ)
X × b = |X||b|sin(θ)
3. The attempt at a solution
I don't know where to start... :( someone pls give me a hint

2. Jan 29, 2015

### Stephen Tashi

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.

3. Jan 30, 2015

### 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?

4. Jan 30, 2015

### Stephen Tashi

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).

5. Jan 30, 2015

### Purcolator

isn't (b x c) = X?

6. Jan 30, 2015

### Stephen Tashi

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.

7. Jan 30, 2015

### 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... :/

8. Jan 30, 2015

### Stephen Tashi

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|.

9. Jan 30, 2015

### Purcolator

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