Find unknown vector X if these relations hold true

  • #1
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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
 

Answers and Replies

  • #2
Stephen Tashi
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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 " [itex] | X \times b | = ....[/itex]

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

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

 
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  • #3
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
Stephen Tashi
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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).
 
  • #5
isn't (b x c) = X?
 
  • #6
Stephen Tashi
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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.
 
  • #7
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
Stephen Tashi
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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|.
 
  • #9
Thanks so much. i finally got the answer... I appreciate you tremendously!!!!!
 

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