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Let be three non-coplanar vectors such that . If , then is equal to ______.
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answer is 36.
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Detailed Solution

We are given three non-coplanar vectors a→, b→, and c→, with the following conditions:
- a→ × b→ = 4c→ (Equation 1)
- b→ × c→ = 9a→ (Equation 2)
- c→ × a→ = αb→, where α > 0 (Equation 3).
The magnitudes of these vectors satisfy:
|a→| + |b→| + |c→| = 36.
Step 1: Orthogonality
Taking the dot product of each cross product equation with the remaining vector:
- From Equation (1), a→ × b→ is perpendicular to both a→ and b→, so a→ ⋅ c→ = 0 and b→ ⋅ c→ = 0.
- Similarly, from Equations (2) and (3), we deduce that a→ ⋅ b→ = 0.
Thus, the vectors a→, b→, and c→ are mutually perpendicular.
Step 2: Relating Magnitudes
From the magnitude of the cross product, we have:
- From Equation (1): |a→ × b→| = |a→||b→| = 4|c→|. (Equation 4)
- From Equation (2): |b→ × c→| = |b→||c→| = 9|a→|. (Equation 5)
- From Equation (3): |c→ × a→| = |c→||a→| = α|b→|. (Equation 6)
Step 3: Combining the Equations
Multiply Equations (4), (5), and (6):
(|a→||b→||c→|)² = (4|c→|)(9|a→|)(α|b→|).
Simplifying,
|a→||b→||c→| = 36α. (Equation 7)
Step 4: Expressing Individual Magnitudes
- From Equation (4): |a→| = (4|c→|) / |b→|.
- From Equation (5): |b→| = (9|a→|) / |c→|.
- From Equation (6): |c→| = (α|b→|) / |a→|.
Using these relations and substituting back into the constraint |a→| + |b→| + |c→| = 36, we solve for α.
Step 5: Final Calculation
From the equations, we find:
- |c→| = 3α,
- |a→| = 2α,
- |b→| = 6.
Substituting into the constraint:
2α + 6 + 3α = 36.
Simplifying:
5α + 6 = 36
5α = 30
α = 6.
Final Answer
The value of α is:
α = 6