Vector Magnitude. |A|

Question

Can I find the magnitude like this?

A=Xi Yj Zk

|A|=sqrt(X Y Z)

Then vector A in polar form is.

A=|A| at angle

Is all this right?

Answer 1

In what follows,

A, i, j, k are vector quantities and

A.A, i.i, i.j are dot products (or scalar products) of 2 vectors

|A|, X, Y, Z are scalar quantities, and

symbol * denotes multiplication of numbers

We should write like this:

A = Xi + Yj + Zk

where i, j, k are unit vectors on 3 axes of coordinates

To evaluate |A|, take the scalar square of both sides:

A.A = (Xi + Yj + Zk).(Xi + Yj + Zk)

Since A.A =|A|*|A|*cos(A,A) = |A|^2*cos0 = |A|^2

and i.i =j.j= k.k = 1*1*cos0 =1

and i.j =j.k =k.i = 1*1*cos(pi/2) = 0

It follows that

|A|^2 =X^2 +Y^2 + Z^2

|A| = sqrt(X^2 +Y^2 + Z^2)

  

Answer 2

You have to square the stuff in the middle: if A = (a, b, c) = a i + b j + c k, then:

|A| = sqrt( a² + b² + c² ).

The vector's "polar form" in 3d will depend on whether you use cylindrical or spherical coordinates.  I encourage you to read up on them in wikipedia:

http://en.wikipedia.org/wiki/Cylindrical...

http://en.wikipedia.org/wiki/Spherical_c...