Naked Science Forum
General Science => General Science => Topic started by: Nishchal on 05/11/2010 18:30:03
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Nishchal asked the Naked Scientists:
Is it possible that the magnitude of the sum and difference of two vextors be equal to each other?
What do you think?
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After a bit of thought I would say the answer is "no", the sum of two vectors cannot be equal to the the difference of the two vectors providing neither of the vectors is of zero magnitude and you are dealing with a Cartesian coordinate system.
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the sum of two vectors cannot be equal to the the difference of the two vectors
But that wasnt the question - the question was Is it possible that the magnitude of the sum and difference of two vextors be equal to each other?
the magnitude of the resultant sum can be the same as a resultant difference - the first example to spring to mind would be if the vectors are perpendicular to each other. Imagine vector A has only horizontal components and vector B has only vertical. The sum (A+B) is a right-triangle and the magnitude is worked out by pythagoras - the difference (A-B) is the same right-triangle flipped over on the horizontal axes. the magnitude is clearly the same even if the resulting vector is not the same. I will do a diagram to explain when I have a bit more time
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Attached diagram - - and, of course, it applies to any perpendicular vectors - not just that are perpendicular to the axes as depicted
[ Invalid Attachment ]
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Aha I see
Tan of 45°
Tan of (45° + 90°)
Tan of (45° + 180°)
Tan of (45° + 90° + 180°)
Magnatudes are the same in different directions
This is restricted to only right triangles
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Nishchal asked the Naked Scientists:
Is it possible that the magnitude of the sum and difference of two vextors be equal to each other?
What do you think?
|u + v| = |u - v| ==> |u + v|2 = |u - v|2
so:
(u + v)•(u + v) = (u - v)•(u - v) where "•" means "scalar product".
Developing the scalar products:
u2 + v2 + 2u•v = u2 + v2 - 2u•v
and so: u•v = 0
That is, the initial equation is valid if and only if the two vectors are orthogonal.
Note that this demonstration is valid in any vectorial space with a defined scalar product, not just R3.
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Of course, I agree that the magnitude can be the same. I didn't read the question carefully enough. D'oh!
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Lightarrow
That's much neater than my algebraic solution
[(ax+bx)2)+(ay+by)2)] - [(ax-bx)2)-(ay-by)2)]=0
Which also ends up with the dot product (axbx + ayby) equals zero.
Perhaps to fully explain - the dot product of two vectors a.b = |a|.|b|CosΘ where |a| is the magnitude and Θ is the angle between the two vectors. For the dot product |a|.|b|CosΘ to equal zero either magnitude of a or b must be zero or CosΘ must be zero - and that happens when Θ is 90 or 270 degrees.
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Lightarrow
That's much neater than my algebraic solution
[(ax+bx)2)+(ay+by)2)] - [(ax-bx)2)+(ay-by)2)]=0
Ive coloured in red the right sign.
Which also ends up with the dot product (axbx + ayby) equals zero.
Perhaps to fully explain - the dot product of two vectors a.b = |a|.|b|CosΘ where |a| is the magnitude and Θ is the angle between the two vectors. For the dot product |a|.|b|CosΘ to equal zero either magnitude of a or b must be zero or CosΘ must be zero - and that happens when Θ is 90 or 270 degrees.
Ok. I didn't explain dot product because, in a general vectorial space, you can define it in a great variety of ways; for example, in quantum mechanics vectors are "functions" Ψ(x,y,z,t) and the dot product is defined as:
(Ψ,Φ) = |Ψ>•|Φ> = ∫∫∫Ψ*(x,y,z,t) Φ(x,y,z,t)dx dy dz
where "Ψ*" means "complex coniugate of Ψ".