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Step-by-Step Solution
Step 1: Interpret the Given Information
We have three real numbers a, b, and c satisfying
$\,a^2 + b^2 + c^2 = 1\,.$
We also know
$$
a \cos(\theta) \;=\; b \cos\!\Bigl(\theta + \tfrac{2\pi}{3}\Bigr) \;=\; c \cos\!\Bigl(\theta + \tfrac{4\pi}{3}\Bigr).
$$
We set θ = $\,\frac{\pi}{9}\,$ (though the specific value of θ ultimately does not change the final angle between the vectors, it is part of the given condition). We want to find the angle between the vectors
$$
\vec{a_1} = a\,\hat{i} + b\,\hat{j} + c\,\hat{k}
\quad \text{and} \quad
\vec{a_2} = b\,\hat{i} + c\,\hat{j} + a\,\hat{k}.
$$
Step 2: Express the Angle Between Two Vectors
The angle $\,\alpha\,$ between two vectors $\,\vec{u}\,$ and $\,\vec{v}\,$ is given by
$$
\cos(\alpha)
= \frac{\vec{u} \cdot \vec{v}}{\lvert \vec{u} \rvert \;\lvert \vec{v} \rvert}.
$$
For
$$
\vec{a_1} = a\,\hat{i} + b\,\hat{j} + c\,\hat{k}
\quad\text{and}\quad
\vec{a_2} = b\,\hat{i} + c\,\hat{j} + a\,\hat{k},
$$
we have:
\[
\vec{a_1}\cdot \vec{a_2} \;=\; ab + bc + ac.
\]
Also,
\[
|\vec{a_1}| = \sqrt{a^2 + b^2 + c^2}
\quad \text{and} \quad
|\vec{a_2}| = \sqrt{a^2 + b^2 + c^2}.
\]
Since $\,a^2 + b^2 + c^2 = 1\,$, it follows
\[
|\vec{a_1}| = |\vec{a_2}| = 1.
\]
Thus,
$$
\cos(\alpha)\;=\;\frac{ab + bc + ac}{1}\;=\;ab + bc + ac.
$$
Step 3: Use the Condition on a, b, and c
We are given that
\[
a \cos(\theta) \;=\; b \cos\!\Bigl(\theta + \tfrac{2\pi}{3}\Bigr) \;=\; c \cos\!\Bigl(\theta + \tfrac{4\pi}{3}\Bigr)
\;=\;\lambda
\quad(\text{say}).
\]
Hence,
\[
a = \frac{\lambda}{\cos(\theta)},
\quad
b = \frac{\lambda}{\cos\Bigl(\theta + \tfrac{2\pi}{3}\Bigr)},
\quad
c = \frac{\lambda}{\cos\Bigl(\theta + \tfrac{4\pi}{3}\Bigr)}.
\]
We wish to find the sum $\,ab + bc + ac\,$ in a convenient form.
Step 4: Analyze the Key Sum Expression
Observe that
\[
\frac{1}{a} \;=\; \frac{\cos(\theta)}{\lambda},
\quad
\frac{1}{b} \;=\; \frac{\cos\!\bigl(\theta + \tfrac{2\pi}{3}\bigr)}{\lambda},
\quad
\frac{1}{c} \;=\; \frac{\cos\!\bigl(\theta + \tfrac{4\pi}{3}\bigr)}{\lambda}.
\]
Thus,
\[
\frac{1}{a} \,+\, \frac{1}{b} \,+\, \frac{1}{c}
\,=\,
\frac{1}{\lambda}
\Bigl[
\cos(\theta)
+
\cos\!\Bigl(\theta + \tfrac{2\pi}{3}\Bigr)
+
\cos\!\Bigl(\theta + \tfrac{4\pi}{3}\Bigr)
\Bigr].
\]
It is a known trigonometric identity that for any $\,\theta\,$,
\[
\cos(\theta)
\;+\;
\cos\!\Bigl(\theta + \tfrac{2\pi}{3}\Bigr)
\;+\;
\cos\!\Bigl(\theta + \tfrac{4\pi}{3}\Bigr)
\;=\; 0.
\]
Therefore,
\[
\frac{1}{a} \,+\, \frac{1}{b} \,+\, \frac{1}{c}
\;=\; 0.
\]
Step 5: Relate This to ab + bc + ac
We can express
\[
ab + bc + ac
\quad \text{in terms of} \quad
a,
\frac{1}{b},
\ldots
\]
But a simpler observation is the following: from the identity
\[
\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 0,
\]
one can argue symmetrically that the product relations lead to
\[
ab + bc + ac \;=\; 0.
\]
(This result is consistent with the trigonometric constraints given for $\,a, b, c\,$.)
Step 6: Compute the Angle
Since
\[
\cos(\alpha)
\;=\;
ab \;+\; bc \;+\; ac
\;=\; 0
\quad\text{(from the analysis above)},
\]
we conclude
\[
\alpha = \frac{\pi}{2}.
\]
Final Answer
Thus, the angle between the two vectors
$\,a\,\hat{i} + b\,\hat{j} + c\,\hat{k}\,$ and
$\,b\,\hat{i} + c\,\hat{j} + a\,\hat{k}\,$ is
$$
\alpha \;=\; \frac{\pi}{2}.
$$
