# Homework Help: Vectors: Find w such that dv/ds = w x v

1. Sep 22, 2012

### YayMathYay

1. The problem statement, all variables and given/known data

T (tangent), B (binormal), N (normal) are an orthogonal triad of unit vectors of a curve in R3.

Given:
• dT/ds = kN
• dN/ds = -kT + tB
• dB/ds = -tN

Find vector w so that these equations may be written in the form:

dv/ds = w x v, where v = T + N + B

2. Relevant equations

Given above!

3. The attempt at a solution

I tried splitting v (T + N + B) into its components, as well as w, and putting them into a matrix to find the determinant (for the cross product).

The matrix consists of <i, j, k>, w = <w1, w2, w3>, and v = <(T1 + N1 + B1), (T2 + N2 + B2), (T3 + N3 + B3)>.

However, taking the determinant/cross product gives mismatching components (ie. j and k components supposedly add up to an i component). I'm not too sure where to go from here :(

Note: This is for a Multivariate Calculus course!

Last edited: Sep 22, 2012
2. Sep 22, 2012

### gabbagabbahey

Hi YayMathYay, welcome to PF!

There's a much easier way to do this than by looking at the Cartesian components of the vectors.

Hint: What is $\mathbf{v}\times ( \mathbf{w} \times \mathbf{v} )$?

3. Sep 22, 2012

### YayMathYay

First of all, thanks for the welcome! :)

I know the hint you gave me means $\left(\mathbf{v} \bullet \mathbf{v}\right)\mathbf{w} - \left(\mathbf{v} \bullet \mathbf{w}\right)\mathbf{v}$. I've tried this method before, but I wasn't too sure where to go from here..

4. Sep 22, 2012

### gabbagabbahey

For starters, what is $\mathbf{v} \cdot \mathbf{v}$ ? Next since $\mathbf{T}$, $\mathbf{B}$ & $\mathbf{N}$ are an orthogonal triad in $\mathbb{R}^3$, they span $\mathbb{R}^3$, and so any vector can be decomposed into a linear combination of those unit vectors. So, why not write $\mathbf{w} = w_{T}\mathbf{T} + w_{N}\mathbf{N} + w_{B}\mathbf{B}$... what does that make $\mathbf{w} ( \mathbf{v} \cdot \mathbf{v} ) - \mathbf{v} ( \mathbf{w} \cdot \mathbf{v} )$?