Q.

Let a, b, c be three non-coplanar vectors such that a×b=4c, b×c=9a and c×a=αb, α>0. If |a|+|b|+|c|=36, then α is equal to ______.

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answer is 36.

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Detailed Solution

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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

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