minwaardekn

Answered

2022-06-24

I'm reading a textbook at the moment that provides the following linear equation,

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

and asks to solve for $\mathbf{v}$. The form of $\mathbf{v}$ is given as

$\mathbf{v}=\frac{{\alpha}^{2}\mathbf{b}-\alpha (\mathbf{b}\times \mathbf{a})+(\mathbf{a}\cdot \mathbf{b})\mathbf{a}}{\alpha ({\alpha}^{2}+|\mathbf{a}{|}^{2})}.$

It's easy enough to verify that this is the correct solution. However, I can't figure out how I'd solve for $\mathbf{v}$ if given just the original equation.

Are there any general approaches to solving this kind of equation systematically?

Edit: $\mathbf{a}$,$\mathbf{b}$ and $\mathbf{v}$ are all vectors, whereas $\alpha $ is a scalar such that $\alpha \ne 0$.

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

and asks to solve for $\mathbf{v}$. The form of $\mathbf{v}$ is given as

$\mathbf{v}=\frac{{\alpha}^{2}\mathbf{b}-\alpha (\mathbf{b}\times \mathbf{a})+(\mathbf{a}\cdot \mathbf{b})\mathbf{a}}{\alpha ({\alpha}^{2}+|\mathbf{a}{|}^{2})}.$

It's easy enough to verify that this is the correct solution. However, I can't figure out how I'd solve for $\mathbf{v}$ if given just the original equation.

Are there any general approaches to solving this kind of equation systematically?

Edit: $\mathbf{a}$,$\mathbf{b}$ and $\mathbf{v}$ are all vectors, whereas $\alpha $ is a scalar such that $\alpha \ne 0$.

Answer & Explanation

Colin Moran

Expert

2022-06-25Added 21 answers

Taking cross product with $\mathbf{a}$ on both sides, we get,

$\begin{array}{rl}& \alpha \mathbf{v}+\mathbf{v}\times \mathbf{a}=\mathbf{b}\\ \phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}& \alpha (\mathbf{v}\times \mathbf{a})+(\mathbf{v}\times \mathbf{a})\times \mathbf{a}=\mathbf{b}\times \mathbf{a}\\ \phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}& \alpha (\mathbf{b}-\alpha \mathbf{v})+(\mathbf{v}\cdot \mathbf{a})\mathbf{a}-|a{|}^{2}\mathbf{v}=\mathbf{b}\times \mathbf{a}\\ \phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}& \alpha \mathbf{b}-{\alpha}^{2}\mathbf{v}+{\displaystyle \frac{1}{\alpha}}(\mathbf{b}\cdot \mathbf{a})\mathbf{a}-|a{|}^{2}\mathbf{v}=\mathbf{b}\times \mathbf{a}& & {\textstyle (}\text{Using}\alpha (\mathbf{v}\cdot \mathbf{a})=\mathbf{b}\cdot \mathbf{a}{\textstyle )}\end{array}$

Now solve for $\mathbf{v}$ directly.

$\begin{array}{rl}& \alpha \mathbf{v}+\mathbf{v}\times \mathbf{a}=\mathbf{b}\\ \phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}& \alpha (\mathbf{v}\times \mathbf{a})+(\mathbf{v}\times \mathbf{a})\times \mathbf{a}=\mathbf{b}\times \mathbf{a}\\ \phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}& \alpha (\mathbf{b}-\alpha \mathbf{v})+(\mathbf{v}\cdot \mathbf{a})\mathbf{a}-|a{|}^{2}\mathbf{v}=\mathbf{b}\times \mathbf{a}\\ \phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}& \alpha \mathbf{b}-{\alpha}^{2}\mathbf{v}+{\displaystyle \frac{1}{\alpha}}(\mathbf{b}\cdot \mathbf{a})\mathbf{a}-|a{|}^{2}\mathbf{v}=\mathbf{b}\times \mathbf{a}& & {\textstyle (}\text{Using}\alpha (\mathbf{v}\cdot \mathbf{a})=\mathbf{b}\cdot \mathbf{a}{\textstyle )}\end{array}$

Now solve for $\mathbf{v}$ directly.

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