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Step-by-Step Solution
Step 1: Interpret the Given Cross-Product Relations
We have three non-coplanar vectors
\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}
with the following conditions:
\overrightarrow{a} \times \overrightarrow{b} = 4\,\overrightarrow{c}
\overrightarrow{b} \times \overrightarrow{c} = 9\,\overrightarrow{a}
\overrightarrow{c} \times \overrightarrow{a} = \alpha\,\overrightarrow{b}, \quad \alpha > 0
|\overrightarrow{a}| + |\overrightarrow{b}| + |\overrightarrow{c}| = \tfrac{1}{36}
Because each cross product is proportional to another single vector,
it implies that each pair of these vectors is mutually perpendicular (their dot products are zero):
\overrightarrow{a} \cdot \overrightarrow{b}
= \overrightarrow{b} \cdot \overrightarrow{c}
= \overrightarrow{c} \cdot \overrightarrow{a} = 0.
Thus,
|\overrightarrow{a} \times \overrightarrow{b}|
= |\overrightarrow{a}||\overrightarrow{b}|\sin 90^\circ
= |\overrightarrow{a}||\overrightarrow{b}|.
Similar interpretations hold for the other cross products.
Step 2: Magnitude Relations from Cross Products
Taking magnitudes on each given cross-product equation yields:
|\overrightarrow{a} \times \overrightarrow{b}| = 4\,|\overrightarrow{c}|
\;\;\Longrightarrow\;\; |\overrightarrow{a}||\overrightarrow{b}| = 4\,|\overrightarrow{c}|.
|\overrightarrow{b} \times \overrightarrow{c}| = 9\,|\overrightarrow{a}|
\;\;\Longrightarrow\;\; |\overrightarrow{b}||\overrightarrow{c}| = 9\,|\overrightarrow{a}|.
|\overrightarrow{c} \times \overrightarrow{a}| = \alpha\,|\overrightarrow{b}|
\;\;\Longrightarrow\;\; |\overrightarrow{c}||\overrightarrow{a}| = \alpha\,|\overrightarrow{b}|.
Let
a = |\overrightarrow{a}|,\; b = |\overrightarrow{b}|,\; c = |\overrightarrow{c}|.
Then the relations become:
a\,b = 4c \quad (1)
b\,c = 9a \quad (2)
c\,a = \alpha\,b \quad (3)
Step 3: Finding a, b, c in Terms of \alpha
Multiply all three relations (1), (2), and (3):
(a\,b)\,(b\,c)\,(c\,a)
\;=\; (4c)\,(9a)\,(\alpha\,b).
The left side is
a^2 b^2 c^2,
and the right side is
4 \times 9 \times \alpha \;(a b c).
Hence,
a^2 b^2 c^2 = 36\,\alpha\,(a\,b\,c)
\;\;\Longrightarrow\;\; a\,b\,c = 36\,\alpha.
\quad (4)
Now divide carefully to solve for each of a, b, c.
Divide (4) by (1):
\dfrac{a\,b\,c}{a\,b} \;=\; \dfrac{36\,\alpha}{4\,c}
\;\;\Longrightarrow\;\; c = \dfrac{36\,\alpha}{4\,c}
\;\;\Longrightarrow\;\; c^2 = 9\,\alpha
\;\;\Longrightarrow\;\; c = 3\sqrt{\alpha}.
Divide (4) by (2):
\dfrac{a\,b\,c}{b\,c} \;=\; \dfrac{36\,\alpha}{9\,a}
\;\;\Longrightarrow\;\; a = \dfrac{36\,\alpha}{9\,a}
\;\;\Longrightarrow\;\; a^2 = 4\,\alpha
\;\;\Longrightarrow\;\; a = 2\sqrt{\alpha}.
Substitute a=2\sqrt{\alpha} and c=3\sqrt{\alpha} back into (3):
c\,a = \alpha\,b
\;\;\Longrightarrow\;\; (3\sqrt{\alpha})(2\sqrt{\alpha}) = \alpha\,b
\;\;\Longrightarrow\;\; 6\,\alpha = \alpha\,b
\;\;\Longrightarrow\;\; b = 6.
Thus
a = 2\sqrt{\alpha}, \; b = 6, \; c = 3\sqrt{\alpha}.
Step 4: Use the Condition on the Sum of Magnitudes
The problem states
|\overrightarrow{a}| + |\overrightarrow{b}| + |\overrightarrow{c}|
= \tfrac{1}{36}.
Substituting
a=2\sqrt{\alpha},\, b=6,\, c=3\sqrt{\alpha},
we get:
(2\sqrt{\alpha}) + 6 + (3\sqrt{\alpha})
\;=\; 5\sqrt{\alpha} + 6
\;=\; \dfrac{1}{36}.
Solving for \sqrt{\alpha} :
5\sqrt{\alpha} + 6 = \dfrac{1}{36}
\;\;\Longrightarrow\;\;
5\sqrt{\alpha} = \dfrac{1}{36} - 6.
Notice that
\dfrac{1}{36} - 6
= \dfrac{1 - 216}{36}
= \dfrac{-215}{36},
which is negative. Hence
5 \sqrt{\alpha}
= \dfrac{-215}{36}
\;\;\Longrightarrow\;\;
\sqrt{\alpha} = \dfrac{-215}{180}
= \text{a negative number}.
Because \sqrt{\alpha} cannot be negative if \alpha > 0,
there is a contradiction.
Step 5: Conclusion
Under the constraint
|\overrightarrow{a}| + |\overrightarrow{b}| + |\overrightarrow{c}|
= \tfrac{1}{36}
and
\alpha > 0,
no real positive value of \alpha satisfies all the given vector relations.
In other words, there is no consistent positive \alpha
that meets the stated conditions as is.
If the problem is exactly as stated, the final outcome is that
the requirement \alpha > 0 cannot be met for the sum of magnitudes
\tfrac{1}{36}.
One either concludes
"No real solution for \alpha "
or that there may be a typographical error
in the given sum of magnitudes.
