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Solve for r if r $×$ b = a $\space$ and given a and b are perpendicular.

My attempt: I already have a method i.e let r = $x$a + $y$b + $z$(a × b) substituting in given equation and comparing the coefficients using the fact that a, b ,a × b are non planar r can be found. But is there a better method to solve this problem. A hint will suffice.

Anton Vrdoljak
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Hari Ramakrishnan Sudhakar
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  • The fact that $\mathrm r \times \mathrm b = \mathrm a$ tells us that either $\mathrm a$ and $\mathrm b$ are perpendicular or $\mathrm a = 0.$ – David K Aug 17 '20 at 14:19
  • In fact, $a$, $b$, $a\times b$ form an orthogonal basis, so there is only one non-zero coefficient to compare. – Miguel Aug 17 '20 at 14:25
  • Are you sure the condition is not supposed to be "$\mathrm r$ and $\mathrm b$ are perpendicular"? – David K Aug 17 '20 at 14:27
  • By the way, when you write an equation, put the whole thing in MathJax, not just individual variables (like $x$). I'm surprised someone bothered to edit this question without fixing that. – David K Aug 17 '20 at 14:42

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i think this is a duplicate : https://math.stackexchange.com/a/32602/427318 seems to give an answer. to explain, if r is perpendicular to b, then it's fairly easy to find the inverse, but any component parallel to b will just disappear due to the bilinearity of the cross product, so there may be an arbitrary nonzero b component (no "a" as it must be perpendicular to both r and b, or 0)

splitting up r into the two components- one parallel to b, and one perpendicular to both a, and b, e.g of the form r= t(bxa)+s(b) and then using the vector triple product identity $\mathbf{a}\times (\mathbf{b}\times \mathbf{c}) = (\mathbf{a}\cdot\mathbf{c})\mathbf{b} -(\mathbf{a}\cdot\mathbf{b}) \mathbf{c}$ + anticommutativity of the cross product to find t should give the answer relatively simply, as most of the terms in the triple product will be zero by orthogonality, thus no need for coefficients

Dylan Davies
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