Vector equation with cross product and unit vector

Question

Does anybody know how to solve the equation

$\mathbf{a} + \mathbf{b} \times \hat{\mathbf{v}} = c \hat{\mathbf{v}},$

where $\mathbf{a}$ and $\mathbf{b}$ are given real vectors, for the unit vector $\hat{\mathbf{v}}$ and the real number $c$?

Answer 1

Write $\bf v$ as a linear combination of $\bf a$, $\bf b$ and $\bf a\times b$. If $\bf a$ and $\bf b$ are linearly independent, you get three independent equations for the unknown coefficients.

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$$\bf v=\alpha {\bf a}+\beta{\bf b}+\gamma {\bf a \times b}$$ Plug it into the equation: $${\bf a}+\alpha {\bf b}\times{\bf a}+\gamma{\bf a}|b|^2-\gamma({\bf a \cdot \bf b}){\bf b}=c(\alpha {\bf a}+\beta{\bf b}+\gamma {\bf a \times b})$$

Equation for the ${\bf a}$ component: $$1+\gamma |b|^2=c\alpha$$ Equation for the ${\bf b}$ component: $$-\gamma ({\bf a \cdot b})=c\beta$$ Equation for the ${\bf a\times b}$ component: $$-\alpha=c\gamma$$ First and last equation combine to give: $$\gamma=-\frac{1}{c^2+|b|^2}$$ $$\alpha=\frac{c}{c^2+|b|^2}$$ The second equation then gives: $$\beta=\frac{({\bf a\cdot \bf b})/c}{c^2+|b|^2}$$

Now, this is a general solution of the above equation. Forcing the vector ${\bf v}$ to have unit length then gives you a condition for $c$. Try to do this yourself. Careful with expansion of all terms in ${\bf v \cdot v}$.

Answer 2

We can write $$ b \times \hat v =[b]_\times \hat v = \pmatrix{0&-b_3&b_2\\b_3&0&-b_1\\-b_2&b_1&0} \hat v $$ Thus, we have $$ \mathbf{a} + \mathbf{b} \times \hat{\mathbf{v}} = c \hat{\mathbf{v}} \implies\\ \mathbf{a} = -[\mathbf{b}]_\times \hat{\mathbf{v}} + cI\; \hat{\mathbf{v}} \implies\\ (cI - [\mathbf{b}]_\times) \hat{\mathbf{v}} = \mathbf{a} \implies\\ \pmatrix{c&b_3&-b_2\\-b_3&c&b_1\\b_2&-b_1&c} \hat {\mathbf v} = \mathbf{a} $$ Happily, this matrix is invertible for all non-zero real $c$.