Let a,b,c R with a^2+b^2+c^2=1.

Question

Let $a,b,c\in\mathbb R$ with $a^2+b^2+c^2=1$. If $a\cos	heta = b\cos\left(	heta+\dfrac{2\pi}3ight) = c\cos\left(	heta+\dfrac{4\pi}3ight)$, where $	heta=\dfrac\pi9$, the angle between $a\hat i+b\hat j+c\hat k$ and $b\hat i+c\hat j+a\hat k$ is:

Accepted Answer

Let the common value be $k$: $a=\dfrac k{\cos\theta}$, etc. Then
$$ab+bc+ca = k^2\left[\frac1{\cos\theta\cos(\theta+\frac{2\pi}3)} + \frac1{\cos(\theta+\frac{2\pi}3)\cos(\theta+\frac{4\pi}3)}+\frac1{\cos(\theta+\frac{4\pi}3)\cos\theta}\right]$$

This sum simplifies (using standard trigonometric identities for angles spaced $\dfrac{2\pi}3$ apart) to exactly $0$, independent of $\theta$.

$$\cos\phi = \frac{(a,b,c)\cdot(b,c,a)}{a^2+b^2+c^2} = ab+bc+ca = 0 \implies \phi=\frac\pi2$$

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