# Thread: Why's that normal = binormal x tangent for a curve parametrized by arc length s?

1. ## Why's that normal = binormal x tangent for a curve parametrized by arc length s?

I'm stuck at the properties of a 3d curve parametrized by arc length. My problem is given below.

We know that the binormal of a curve at a point is:

$\displaystyle b = n \times t$

where $\displaystyle b =$ binormal and $\displaystyle n$ is normal to the curve and $\displaystyle t$ is tangent.

So binormal is the cross product of normal and tangent at a point of a curve.

Now if $\displaystyle b = n \times t$ why's that $\displaystyle n = b \times t$ ? How do one prove this statement?

What properties of mathematics verify that $\displaystyle n = b \times t$? Is it possible to kindly help me answer this question?

2. ## Re: Why's that normal = binormal x tangent for a curve parametrized by arc length s?

With $\displaystyle b= n\times t$, there is NO "property of mathematics" that gives "$\displaystyle n= b\times t$". It is, rather, true that $\displaystyle n= t\times b$.

n and t are also unit vectors. Further, the length of $\displaystyle u\times v$ is $\displaystyle |u||v|sin(\theta)$ where $\displaystyle \theta$ is the angle between the two vectors. Since n and t are at right angles, b is also a unit vector, perpendicular to both n and t. That is, we have the "cyclic" formulas, $\displaystyle b= n\times t$, $\displaystyle n= t\times b$ and $\displaystyle t= b\times n$.

3. ## Re: Why's that normal = binormal x tangent for a curve parametrized by arc length s?

Thanks HallsofIvy.

Then if $\displaystyle a = b \times c$ for any vector $\displaystyle a, b$ and $\displaystyle c$ then can we say $\displaystyle b = a \times c$ and $\displaystyle c = a \times b$?

4. ## Re: Why's that normal = binormal x tangent for a curve parametrized by arc length s?

x3bnm said:
Then if $\displaystyle a = b \times c$ for any vector $\displaystyle a, b$ and $\displaystyle c$ then can we say $\displaystyle b = a \times c \text{ and } c = a \times b$?

I got it. My last statement is wrong. For those who want to understand: the Counterexample to my statement:

Suppose $\displaystyle b = (1,2,1), c = (2, 3, 1) \text{ and } a = b \times c = (-1, 1, -1)$

But $\displaystyle b \neq a \times c \text{ and } c \neq a \times b$

The statement that $\displaystyle n = b \times t$ is true for binormal, normal and tangent.