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Show that this set of equations defines a unique vector

  • Thread starter 1MileCrash
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  • #1
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Homework Statement



Show that, given vectors A and B:

A * X = |a|
and
A x X = B

defines X as a unique vector

Homework Equations





The Attempt at a Solution



No idea.

My only conclusion thus far is that the magnitude of x is equal to 1/cos(theta) where theta is the angle it makes with vector A.

Any ideas?
 

Answers and Replies

  • #2
SammyS
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Homework Statement



Show that, given vectors A and B:

A * X = |a|
and
A x X = B

defines X as a unique vector

Homework Equations



The Attempt at a Solution



No idea.

My only conclusion thus far is that the magnitude of x is equal to 1/cos(theta) where theta is the angle it makes with vector A.

Any ideas?
It appears that you're being a bit careless with upper/lower case.

If you mean AX = |A|,

then, yes, the magnitude of vector, X, is given by |X| = 1/cos(θ) .

******************

So, it looks like you used CD = |C| |D| cos(θ) .

Now use that the magnitude of the vector product is:
|C × D| = |C| |D| sin(θ)

 
  • #3
1,331
45
Sorry! Yes, I tried to make all vectors upper case afterwards to prevent confusion with the cross product.

I have also noted the magnitude of the cross product to be |A||X|sin(theta) = |B| but I am not sure how this relates to finding anything out about X, since I already know its magnitude based on theta above, and I know nothing about B other than it being perpendicular to both A and X.
 
  • #4
SammyS
Staff Emeritus
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Homework Helper
Gold Member
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Sorry! Yes, I tried to make all vectors upper case afterwards to prevent confusion with the cross product.

I have also noted the magnitude of the cross product to be |A||X|sin(theta) = |B| but I am not sure how this relates to finding anything out about X, since I already know its magnitude based on theta above, and I know nothing about B other than it being perpendicular to both A and X.
And what is |X| ?
 
  • #5
1,331
45
Ah!

So then that results in
Tan(theta) = |B|/|A|

So it's direction relative to A is
arctan(|B|/|A|)

And it's magnitude is
1/cos(theta)
Which is of course:

1/cos(arctan(|B|/|A|)

Geometrically, I figure that cos(arctan(x)) = (x^2 + 1)^(-1/2)
So 1/cos(arctan(x)) = (x^2 + 1)^(1/2)

Then letting x = |B|/|A|,

The magnitude must be
[itex]\sqrt{\frac{|B|^{2}}{|A|^{2}} + 1}[/itex]

And its direction relative to A must be
[itex]arctan(\frac{|B|}{|A|})[/itex]


Hm... but arctan will give the same thing for some angle theta, and 180 - theta. So it's still not unique.. this must be resolved some way?


Thanks for the help!
 

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