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What I've got is this:

$$ (a_y b_y + a_z b_z) \hat{i} + (a_x b_x + a_z b_z) \hat{j} + (a_x b_x + a_y b_y) \hat{k} $$

That looks like a cross product, but it isn't a cross product. Is there any name for this?

Additionally:

This came up in solving the calculus of variations problem of a rocket in flight in the atmosphere under a very simple drag formulation. So with the Hamiltonian as a function of the state, costate and controls, ignoring the mass costate and other details:

$$ H(\vec{R},\vec{V},\vec{P_v},\vec{P_r},\vec{u}) = \vec{P_r} \cdot \vec{V} + \vec{P_v} \cdot ( - \frac{\vec{R}}{\lvert R \rvert ^3} + T * \vec{u} - \frac{D(\vec{R},\vec{V})}{m}) $$

And with drag as function of R and V:

$$ \begin{align} h &= \lvert R \rvert - R_{body} \\ \rho &= \rho_0* e^{-h/h_{scale}} \\ C1 &= 0.5 * A_{ref}*C_d * \rho * \lvert V \rvert^2 \\ D(\vec{R},\vec{V}) &= C1 * \frac{V}{\lvert V \rvert} \\ \end{align} $$

Where $T, R_{body}, h_{scale}, \rho_0, A_{ref}, C_d$ are all constant.

And one of the costate equations is:

$$ \frac{\partial H}{\partial \vec{V}} = - \dot{P_v} $$

I have the solution to that equation from Mathematica, but it is rather too long to write out in MathJax, and I'm trying to condense it (but using vector operations and not just the component-wise equation that Mathematica spits out).

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    $\begingroup$ Where did you come across this? $\endgroup$ Dec 18, 2022 at 7:44
  • $\begingroup$ Is $1,i,j,k$ the basis for the quaternion algebra? $\endgroup$ Dec 18, 2022 at 8:59
  • $\begingroup$ cartesian x,y,z basis vectors $\endgroup$
    – lamont
    Dec 18, 2022 at 20:26

1 Answer 1

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Well I don't know what it is called, but I figured out how to express it as vector operations. I probably should have given more context:

$$ b_x (a_y b_y + a_z b_z) \hat{i} + b_y (a_x b_x + a_z b_z) \hat{j} + b_z (a_x b_x + a_y b_y) \hat{k} $$

This is equal to:

$$ b_x (a_x b_x + a_y b_y + a_z b_z - a_x b_x) \hat{i} + b_y (a_x b_x + a_y b_y + a_z b_z - a_y b_y) \hat{j} + b_z (a_x b_x + a_y b_y + a_z b_z - a_z b_z) \hat{k} $$

I can now see the dot product in there:

$$ \vec{b} ( \vec{a} \cdot \vec{b} ) - \vec{b} * \vec{a} * \vec{b} $$

Where I'm using $*$ to mean the element-wise Hadamard product: $\vec{a} * \vec{b} = a_x b_x \hat{i} + a_y b_y \hat{j} + a_z b_z \hat{k}$

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