Let a→, b→, c→ be three non-coplanar vectors such that a→×b→=4c→, b→×c→=9a→ and c→×a→=αb→, α0. If |a→|+|b→|+|c→|=36, then α is equal to ______.

Question

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 ______.

Accepted Answer

We are given three non-coplanar vectors a→, b→, and c→, with the following conditions:

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:

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

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:

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

Let $\vec{a}, \vec{b}, \vec{c}$ be three non-coplanar vectors such that $\vec{a} \times \vec{b} = 4 \vec{c}, \vec{b} \times \vec{c} = 9 \vec{a} and \vec{c} \times \vec{a} = α \vec{b}, α > 0$ . If $| \vec{a} | + | \vec{b} | + | \vec{c} | = 36$ , then $α$ is equal to ______.