Solving Dot & Cross Product Equations: Vector Positioning

[kingwinner]

Q:Given non-zero vectors u and v, find a vector x such that both equalitites x dot u=|u|, and x cross u = v hold. What should be the mutual position of the given vectors u and v in space, so that the problem has a solution?


What I've got so far:

Setting x = au + b(u cross v)
I have that
a=1/|u| and b= +/- |v| / |u cross (u cross v)|

This answers the first part of the question, but I don't get the second part (i.e. What should be the mutual position of the given vectors u and v in space, so that the problem has a solution?)

Thanks for helping and explaining!

 

[kingwinner]

This question looks hard and weird, but I am sure someone here knows the answer...

 

[HallsofIvy]

Q:Given non-zero vectors u and v, find a vector x such that both equalitites x dot u=|u|, and x cross u = v hold. What should be the mutual position of the given vectors u and v in space, so that the problem has a solution?


What I've got so far:

Setting x = au + b(u cross v)
I have that
a=1/|u| and b= +/- |v| / |u cross (u cross v)|

This answers the first part of the question, but I don't get the second part (i.e. What should be the mutual position of the given vectors u and v in space, so that the problem has a solution?)

Thanks for helping and explaining!

I started to say this was impossible until I saw that second question! Just "any" u and v won't do. You know, I expect, that the cross product of two vectors is perpendicular to both. Okay, if x cross u= v, what must be true of u and v?

 

[nizi]

using the second equation,
[tex] \left( { x \times u } \right) \times u = v \times u [/tex]
①[tex] u \left( { x \cdot u } \right) - x \left( { u \cdot u } \right) = v \times u [/tex]
using the first equation and developping
②[tex] x = \frac{1}{\left| u \right|} u - \frac{1}{\left| u \right|^{2}} v \times u [/tex]
hence x is obtained.
then considering the following 3 eqations,
[tex] \left( { x \times u } \right) \cdot u = v \cdot u [/tex]
[tex] \left( { x \times u } \right) \times v = v \times v [/tex]
③[tex] \left( { x \times u } \right) \cdot v = v \cdot v [/tex]
respectively
④[tex] 0 = v \cdot u [/tex]
⑤[tex] u \left( { x \cdot v } \right) - x \left( { v \cdot u } \right) = 0 [/tex]
l.h.s in ③ [tex] = x \cdot \left( { u \times v } \right) [/tex] so substituting ① into ③
⑥[tex] x \cdot \left( { - u \left( { x \cdot u } \right) + x \left( { u \cdot u } \right) } \right) = \left| v \right|^{2} [/tex]
substituting ④ into ⑤
[tex] u \left( { x \cdot v } \right) = 0 [/tex]
i.e
⑦[tex] x \cdot v = 0 [/tex] because [tex] u \neq 0 [/tex]
according to ④ and ⑦ we can say [tex] u [/tex] and [tex] v [/tex] are orthogonal and [tex] u [/tex] and [tex] x [/tex] are on the same plane perpendicular to [tex] v [/tex] and not parallel due to the given first equation.
further using ①, ⑥ we can also say the relationship of the norm of [tex] u [/tex], [tex] v [/tex] and [tex] x [/tex] as
[tex] \left| u \right|^{2} \left( { \left| x \right|^{2} - 1 } \right) = \left| v \right|^{2} [/tex]

 

[kingwinner]

I forgot to type the hint that is attached to the orginal question:
HINT: seek the vector x in the form of a linear combination of u and (u cross v)

But I'm not sure how this will help...

 

[kingwinner]

Okay, is x cross u= v, what must be true of u and v?

So v and u must be perpendicular, right?
Is that all?

 

[HallsofIvy]

Yes.

 

[kingwinner]

Yes.

But there is another condition that it must satisfy: x dot u=|u|, does this impose any other restrictions?