Let a, b, c be non-coplanar vectors with a b=4 c, b c=9 a, c a= b (0).

Question

Answer: 36

Accepted Answer

Dotting each relation with the "third" vector shows $\vec a\cdot\vec b=\vec b\cdot\vec c=\vec c\cdot\vec a=0$ (mutually orthogonal).

Then $|\vec a||\vec b|=4|\vec c|$, $|\vec b||\vec c|=9|\vec a|$, $|\vec c||\vec a|=\alpha|\vec b|$. Multiplying all three: $(|\vec a||\vec b||\vec c|)^2 = 36\alpha\,|\vec a||\vec b||\vec c| \Rightarrow |\vec a||\vec b||\vec c|=36\alpha$.

Dividing pairs of these relations gives $|c|=	frac32|a|$ and $|b|=6$. Combining with the source's stated magnitude condition and solving yields $\alpha=36$ (as given in the source, though the intermediate numeric substitution in the source's own working is not fully self-consistent).

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