Prove. The vector dB/ds is perpendicular to vectors T and B and thus is some multiple of vector N.

Proof. Since the vectors T, N and B are mutually perpendicular unit vectors we have

1) B∙T = 0.

Taking the derivative of both sides of 1) we get

2) (dB/ds)∙T + B∙(dT/ds) = 0 .

Now dT/ds = κN so 2) becomes

3) (dB/ds)∙T + κB∙N = 0 .

But B∙N = 0 so 3) becomes

4) (dB/ds)∙T = 0

which proves that dB/ds is perpendicular to T.

We now prove that dB/ds is perpendicular to B.

First we note that

5) B∙B = 0 .

Taking the derivative of both sides of 5) we get

6) (dB/ds)∙B + B∙(dB/ds) = 0 .

or

7) 2 B∙(dB/ds) = 0 .

Thus B∙(dB/ds) = 0 and dB/ds is consequently perpendicular to B.