If I have 2 vectors A and B in 3D,how do I find the normal component and parallel component of one on another??

Suppose we wish to write A in terms of components that are normal to B, A_n, and parallel to B, A_p.

A=A_p+A_n

The magnitude of the parallel component will simply be the dot product of the two vectors.

|A_p| = A\cdot B

Knowing its magnitude, we can then write it as a vector by multiplying it by a unit vector in the direction of B.

A_p = |A_p|\frac{B}{|B|}

You can then find the normal component by plugging back into the first equation above:

A_n=A-A_p

Hopefully you are familiar with the dot product. If not, try googling it; there should be thousands of references out there. If you still need help, just ask.

Thanks a lot.. one more question, if I need to find the normal to a given unit vector, how do I go about it..

The unit normal vector can be figured by using

N(t)=\frac{T'(t)}{||T'(t)||}

Let's do an example.

Say we have x=a\cdot cos(t), \'\ y=a\cdot sin(t), \;\ z=ct, \;\ a>0

This is a circular helix.

The radius vector is r(t)=(a\cdot cos(t))i+(a\cdot sin(t))j+(ct)k

r'(t)=(-a\cdot sin(t))i+(a\cdot cos(t))j+ck

||r'(t)||=\sqrt{(-a\cdot sin(t))^{2}+(a\cdot cos(t))^{2}+c^{2}}=\sqrt{a^{2}+c^{2}}

T(t)=\frac{r'(t)}{||r'(t)||}=\frac{-a\cdot sin(t)}{\sqrt{a^{2}+c^{2}}}i+\frac{a\cdot cos(t)}{\sqrt{a^{2}+c^{2}}}j+\frac{c}{\sqrt{a^{2}+ c^{2}}}k

T'(t)=\frac{-a\cdot cos(t)}{\sqrt{a^{2}+c^{2}}}i-\frac{a\cdot sin(t)}{\sqrt{a^{2}+c^{2}}}j

||T'(t)||=\frac{a}{\sqrt{a^{2}+c^{2}}}

\fbox{N(t)=\frac{T(t)}{||T'(t)||}=(-cos(t))i-(sin(t))j}

Because the k component of N(t) is 0, this vector line lies in a horizontal plane for every value of t.

N(t) points toward the z-axis for all t.

Does that help? I hope I addressed what you are asking for.