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Step-by-Step Solution
Step 1: List the Given Information
We have three vectors:
1. |\vec{a}| = \sqrt{31} .
2. 4|\vec{b}| = |\vec{c}| = 2 \implies |\vec{b}| = \frac{1}{2}, \;|\vec{c}| = 2 .
3. 2(\vec{a} \times \vec{b}) = 3(\vec{c} \times \vec{a}) .
4. The angle between \vec{b} and \vec{c} is \frac{2\pi}{3} (i.e., 120^\circ ).
Step 2: Use the Cross-Product Condition
From
2(\vec{a} \times \vec{b}) = 3(\vec{c} \times \vec{a}),
we can rewrite as
\vec{a} \times (2\vec{b} + 3\vec{c}) = \vec{0}.
This implies that \vec{a} is parallel to (2\vec{b} + 3\vec{c}) . Therefore, we can write
\vec{a} = \lambda \bigl(2\vec{b} + 3\vec{c}\bigr)
for some real scalar \lambda .
Step 3: Determine \lambda Using the Magnitudes
Since |\vec{a}| = \sqrt{31} and \vec{a} = \lambda(2\vec{b} + 3\vec{c}) , we have:
|\vec{a}|^2
= \lambda^2 \,\Bigl[4|\vec{b}|^2 \;+\; 9|\vec{c}|^2 \;+\; 12\,(\vec{b}\cdot \vec{c})\Bigr].
• |\vec{b}| = \tfrac{1}{2} \implies |\vec{b}|^2 = \tfrac{1}{4}.
• |\vec{c}| = 2 \implies |\vec{c}|^2 = 4.
• \vec{b}\cdot \vec{c} = |\vec{b}||\vec{c}|\cos\bigl(\tfrac{2\pi}{3}\bigr)
= \tfrac{1}{2} \times 2 \times \bigl(-\tfrac{1}{2}\bigr) = -\tfrac{1}{2}.
Substituting these into the bracket:
4|\vec{b}|^2 + 9|\vec{c}|^2 + 12(\vec{b}\cdot \vec{c})
= 4\Bigl(\tfrac{1}{4}\Bigr) + 9(4) + 12\Bigl(-\tfrac{1}{2}\Bigr)
= 1 + 36 - 6 = 31.
Thus
|\vec{a}|^2
= \lambda^2 \times 31
= 31
\quad \Longrightarrow \quad \lambda^2 = 1
\quad \Longrightarrow \quad \lambda = \pm 1.
Hence,
\vec{a} = \pm \bigl(2\vec{b} + 3\vec{c}\bigr).
Step 4: Express \vec{a}\times \vec{c} and \vec{a}\cdot \vec{b}
Taking \vec{a} = \pm (2\vec{b} + 3\vec{c}) :
1) \vec{a}\times \vec{c}
= \pm\bigl[(2\vec{b} + 3\vec{c})\times \vec{c}\bigr]
= \pm\Bigl[2(\vec{b}\times \vec{c}) + 3(\vec{c}\times \vec{c})\Bigr]
= \pm\,2(\vec{b}\times \vec{c}),
since \vec{c}\times \vec{c} = \vec{0}.
2) \vec{a}\cdot \vec{b}
= \pm\bigl[(2\vec{b} + 3\vec{c})\cdot \vec{b}\bigr]
= \pm\Bigl[2|\vec{b}|^2 + 3(\vec{c}\cdot \vec{b})\Bigr].
We know |\vec{b}|^2 = \tfrac{1}{4} and \vec{c}\cdot \vec{b} = -\tfrac{1}{2} . Thus
2|\vec{b}|^2 = 2\times \tfrac{1}{4} = \tfrac{1}{2},
\quad 3(\vec{c}\cdot \vec{b}) = 3\times\bigl(-\tfrac{1}{2}\bigr) = -\tfrac{3}{2}.
Hence,
2|\vec{b}|^2 + 3(\vec{c}\cdot \vec{b})
= \tfrac{1}{2} - \tfrac{3}{2}
= -1.
This shows |\vec{a}\cdot \vec{b}| = 1.
Step 5: Find |\vec{b}\times \vec{c}|
The magnitude of the cross product is
|\vec{b} \times \vec{c}|
= |\vec{b}|\;|\vec{c}|\;\sin\bigl(\tfrac{2\pi}{3}\bigr)
= \Bigl(\tfrac{1}{2}\Bigr)\times 2 \times \tfrac{\sqrt{3}}{2}
= \tfrac{\sqrt{3}}{2}.
Step 6: Compute the Desired Expression
Since \vec{a}\times \vec{c} = \pm\,2(\vec{b}\times \vec{c}) and |\vec{a}\cdot \vec{b}| = 1 , we have
\frac{|\vec{a}\times \vec{c}|}{|\vec{a}\cdot \vec{b}|}
= 2\,|\vec{b}\times \vec{c}|
= 2 \times \tfrac{\sqrt{3}}{2}
= \sqrt{3}.
Squaring this quantity:
\Bigl(\frac{\vec{a}\times \vec{c}}{\vec{a}\cdot \vec{b}}\Bigr)^2 = (\sqrt{3})^2 = 3.
Final Answer
\boxed{3}
