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Step-by-Step Solution
Step 1: Understand the Given Conditions
We have two vectors $ \overrightarrow{a} $ and $ \overrightarrow{b} $ satisfying:
1) $ \lvert 2\overrightarrow{a} + 3\overrightarrow{b} \rvert = \lvert 3\overrightarrow{a} + \overrightarrow{b} \rvert $
2) The angle between $ \overrightarrow{a} $ and $ \overrightarrow{b} $ is $ 60^\circ $.
3) $ \tfrac{1}{8}\overrightarrow{a} $ is a unit vector, hence $ \lvert \overrightarrow{a} \rvert = 8 $.
Step 2: Square Both Sides of the Magnitude Equation
Since $ \lvert 2\overrightarrow{a} + 3\overrightarrow{b} \rvert = \lvert 3\overrightarrow{a} + \overrightarrow{b} \rvert $, we square each side:
$$
\lvert 2\overrightarrow{a} + 3\overrightarrow{b} \rvert^2 = \lvert 3\overrightarrow{a} + \overrightarrow{b} \rvert^2.
$$
In dot product form:
$$
(2\overrightarrow{a} + 3\overrightarrow{b}) \cdot (2\overrightarrow{a} + 3\overrightarrow{b}) \;=\; (3\overrightarrow{a} + \overrightarrow{b}) \cdot (3\overrightarrow{a} + \overrightarrow{b}).
$$
Step 3: Expand the Dot Products
Expand both sides:
Left-hand side:
$$
(2\overrightarrow{a} + 3\overrightarrow{b}) \cdot (2\overrightarrow{a} + 3\overrightarrow{b})
= 4\,\lvert \overrightarrow{a} \rvert^2
+ 12\,(\overrightarrow{a} \cdot \overrightarrow{b})
+ 9\,\lvert \overrightarrow{b} \rvert^2.
$$
Right-hand side:
$$
(3\overrightarrow{a} + \overrightarrow{b}) \cdot (3\overrightarrow{a} + \overrightarrow{b})
= 9\,\lvert \overrightarrow{a} \rvert^2
+ 6\,(\overrightarrow{a} \cdot \overrightarrow{b})
+ \lvert \overrightarrow{b} \rvert^2.
$$
Therefore, setting them equal gives:
$$
4\,\lvert \overrightarrow{a} \rvert^2 + 12\,(\overrightarrow{a} \cdot \overrightarrow{b}) + 9\,\lvert \overrightarrow{b} \rvert^2
=
9\,\lvert \overrightarrow{a} \rvert^2 + 6\,(\overrightarrow{a} \cdot \overrightarrow{b}) + \lvert \overrightarrow{b} \rvert^2.
$$
Step 4: Rearrange Terms
Subtracting common terms, we get:
$$
5\,\lvert \overrightarrow{a} \rvert^2 - 6\,(\overrightarrow{a} \cdot \overrightarrow{b})
= 8\,\lvert \overrightarrow{b} \rvert^2.
$$
Step 5: Use the Relationship Between $ \overrightarrow{a} $ and $ \overrightarrow{b} $
We know $ \lvert \overrightarrow{a} \rvert = 8 $ and the angle between $ \overrightarrow{a} $ and $ \overrightarrow{b} $ is $ 60^\circ $. Hence,
$$
\overrightarrow{a} \cdot \overrightarrow{b}
= \lvert \overrightarrow{a} \rvert \,\lvert \overrightarrow{b} \rvert \cos(60^\circ)
= 8\,\lvert \overrightarrow{b} \rvert \times \tfrac{1}{2}
= 4\,\lvert \overrightarrow{b} \rvert.
$$
Substitute these into the equation:
$$
5 \times (8)^2 - 6 \times 4\,\lvert \overrightarrow{b} \rvert
= 8\,\lvert \overrightarrow{b} \rvert^2.
$$
Step 6: Solve for $ \lvert \overrightarrow{b} \rvert $
Substitute $ \lvert \overrightarrow{a} \rvert = 8 $:
$$
5 \times 64 \;-\; 24\,\lvert \overrightarrow{b} \rvert
= 8\,\lvert \overrightarrow{b} \rvert^2.
$$
Simplify:
$$
320 - 24\,\lvert \overrightarrow{b} \rvert
= 8\,\lvert \overrightarrow{b} \rvert^2.
$$
Move all terms to one side:
$$
8\,\lvert \overrightarrow{b} \rvert^2 + 24\,\lvert \overrightarrow{b} \rvert - 320 = 0.
$$
Divide by 8:
$$
\lvert \overrightarrow{b} \rvert^2 + 3\,\lvert \overrightarrow{b} \rvert - 40 = 0.
$$
Step 7: Factorize and Find the Positive Root
Factor the quadratic:
$$
(\lvert \overrightarrow{b} \rvert + 8)\,(\lvert \overrightarrow{b} \rvert - 5) = 0.
$$
Hence, the possible solutions are:
$$
\lvert \overrightarrow{b} \rvert = -8 \quad \text{(reject negative magnitude)}
\quad \text{or} \quad
\lvert \overrightarrow{b} \rvert = 5.
$$
Since magnitude must be positive, we accept:
$$
\lvert \overrightarrow{b} \rvert = 5.
$$
Final Answer:
$ \lvert \overrightarrow{b} \rvert = 5.
