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debjit625

Member

Registered: 2012-07-23

Posts: 101

Square Root of a square of a cross product

Hi I got a problem

Let a and b be two vectors...
(a x b) = |a| |b| sin(t) n

where  t is the angle between the two vectors a and b ,and  n is the unit vector in the direction as per right hand rule.

If I square the equation
(a x b)^2 = |a|^2 |b|^2 (sin(t))^2 n^2

(a x b)^2 = |a|^2 |b|^2 (sin(t))^2

Now if I take the square root of both the sides

(a x b) = |a| |b| sin(t)  ...which is the problem LHS is vector and RHS is scaler

I think there is some rules for vectors which I didn't follow may be... but what please explain and give me some reference where I can read more about this.

Thanks

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Debjit Roy
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The essence of mathematics lies in its freedom - Georg Cantor

zetafunc

Moderator

Registered: 2014-05-21

Posts: 2,436

Website

Re: Square Root of a square of a cross product

I assume by 'squaring' you mean the dot product with itself, i.e. (a x b)^2 = (a x b).(a x b)?

When you square root (a x b)^2, you don't get (a x b). Instead, you get the scalar quantity, |a x b|.

debjit625

Member

Registered: 2012-07-23

Posts: 101

Re: Square Root of a square of a cross product

OK I got that (It should be the case), but I want to know if that is a vector rule? if yes can you give me a resource where I can understand why its like that and more.

zetafunc Thanks

Last edited by debjit625 (2015-12-10 20:40:24)

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Debjit Roy
___________________________________________________
The essence of mathematics lies in its freedom - Georg Cantor

zetafunc

Moderator

Registered: 2014-05-21

Posts: 2,436

Website

Re: Square Root of a square of a cross product

MathsIsFun has a basic page on vectors here: https://www.mathsisfun.com/algebra/vectors.html

This page seems a little more involved.

I don't know if

   is necessarily a 'vector rule' as such, but rather it is how the square root function is defined. For any real number x, we have

. In this case, the quantity

is no longer a vector, it is a scalar (a real number). So it makes sense that we can treat it like one.

Michael Spivak's Calculus on Manifolds gives a more sophisticated viewpoint of the cross product, but is consistent with its usual definition, and shows how the cross product fits in with the other concepts discussed in that text (in this case, an alternating 1-tensor). But I don't recommend going through that chapter without a first course in linear algebra.

Last edited by zetafunc (2015-12-10 21:10:57)

debjit625

Member

Registered: 2012-07-23

Posts: 101

Re: Square Root of a square of a cross product

Thanks

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Debjit Roy
___________________________________________________
The essence of mathematics lies in its freedom - Georg Cantor