Why is A.A = ||A||^2 in Vector Analysis?

[parshyaa]

Why A .A =||A||^2(dot product) in vector analysis. Every where in vector analysis mathematicians used this result. To prove A.B = ||A|| ||B||cos(theta) we assume that A.A is ||A||^2 , without assuming this we can't prove A.B = ||A|| ||B||cos(theta) . I think they assumed it because dot product is approximately a multiplication of magnitude (only for A.A because both vectors have same direction therefore it will not effect the direction of resultant and in dot product we don't need direction, therefore we simply write A.A= ||A||^2)

• I want to know your opinion on this, is there a good reason for this.
• And please do not say that A.A = ||A||^2 because cos(theta) =1 in this case.

 

[ShayanJ]

Its just a definition! People define the norm of a vector to be the square root of the inner product of that vector with itself. Its called a norm induced by the inner product. You can define other norms on a vector space but its a natural definition for an inner product space.

 

[parshyaa]

Okk Then why A cross A is 0

Its just a definition! People define the norm of a vector to be the square root of the inner product of that vector with itself. Its called a norm induced by the inner product. You can define other norms on a vector space but its a natural definition for an inner product space.

cross

 

[ShayanJ]

Okk Then why A cross A is 0

cross

I don't see how the two questions can be related but I try to answer.

Whatever you want to do with a mathematical operation, you should start from defining it.

One definition is $\vec A \times \vec B=\left| \begin{array}{ccc}\hat x \ \ \ \ \ \ \hat y \ \ \ \ \ \ \hat z \\ A_x \ \ \ A_y \ \ \ A_z \\ B_x \ \ \ B_y \ \ \ B_z \end{array} \right|$ where $| |$ means calculating the determinant. And we know that when one row(column) of a determinant is a multiple of another row(column), the determinant is equal to zero. So by this definition, any two vectors that are parallel, have zero cross product.

Another definition is $\vec A \times \vec B=AB \sin \phi \hat n$ where $\phi$ is the angle between the two vectors and $\hat n$ is a unit vector perpendicular to the plane containing the two vectors and in a direction given by the right-hand-rule. Then its easily seen that whenever $\phi=0$, this gives zero. So again, two parallel vectors give zero cross product.

Another definition is by defining the cross product of two vectors, as a vector with the magnitude equal to the area of the parallelogram that those two vectors form together which is along the normal vector of that parallelogram and its direction is given by the right-hand-rule. Two parallel vectors can't form a parallelogram and again we see that the cross product of two parallel vectors is zero.

I'm not sure about this one, but I think you even can define it as a product between vectors that gives a vector and has the property $\vec A\times \vec A=0$. So if this definition is equivalent to other definitions, then $\vec A\times \vec A=0$ can even be a defining property!

So I think your question boils down to "why should we use one of the above definitions?". Easy...because this is the operation that is useful to us and so this is the operation that we need to give a name to so we can refer to it easily. If anyone feels the need to define an operation between vectors that gives a non-zero result for two parallel vectors, Its OK, he can define it. But that will be another operation with another name that is useful for other situations!

 

[parshyaa]

[QUOTE="Shayan.J, post: 5529094, member:

Another definition is by defining the cross product of two vectors, as a vector with the magnitude equal to the area of the parallelogram that those two vectors form together which is along the normal vector of that parallelogram and its direction is given by the right-hand-rule. Two parallel vectors can't form a parallelogram and again we see that the cross product of two parallel vectors is zero.I liked this definition, you want to say that we can define any type of operation and if it holds well we can use it and name it. Thanks man you are awesome.

 

[ShayanJ]

you are awesome.

Yeah, I know!

 

[parshyaa]

How will you define dot product

 

[ShayanJ]

See here!