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radiation · Sep 3, 2009, 08:54 PM

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

jcaron2 · Sep 4, 2009, 11:46 AM

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.

radiation · Sep 5, 2009, 10:54 PM

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

galactus · Sep 6, 2009, 04:23 AM

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.