I have an equation: a=b xx c, where a b and c are 3-vectors. How could I derive b from the equation and express it in terms of a and c?

Arectemieryf0

Arectemieryf0

Answered question

2022-07-19

Derive b from a = b × c
I have an equation:
a = b × c ,
where a b and c are 3-vectors.
How could I derive b from the equation and express it in terms of a and c ?

Answer & Explanation

Quenchingof

Quenchingof

Beginner2022-07-20Added 14 answers

You can't. Given c and a vector a perpendicular perpendicular to c choose b in the plane perpendicular to a. Then b × c lies in the direction of a and you want | a | = | b | | c | sin θ where θ is the angle between b and c taken in the correct direction.
So then all you need is | b | sin θ = | a | | c | . If you choose θ then the length of b is fixed, but you can choose any non-zero θ, so b is not unique and can't be determined
PoentWeptgj

PoentWeptgj

Beginner2022-07-21Added 6 answers

Note that
if c a 0 then there are no solutions.
if c = 0 , then there are no solutions unless a=0, in which case every b works.
if c 0 , then p × c = 0 if and only if p is a scalar multiple of c.
So if a, a , c 0 and c a = 0, the solutions are b = b 0 + λ c , some b 0 such that b 0 × c = a
How could we find such a b 0 ? One way is to select it so that b 2 is minimized. That is, b 0 is perpendicular to c . Since b 0 is perpendicular to a (from a = b × c ), we get b 0 = μ c × a . To solve for μ, we take scalar triple product
b ( c × a ) = a ( b × c ) = a a = a 2
so
μ = a 2 | c × a | 2
and so the solutions are b = a 2 | c × a | 2 ( c × a ) + λ c for some scalar λ

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