Understanding Coplanar Vectors

Vectors are said to be coplanar if they lie in the same plane. A fundamental condition for three vectors to be coplanar is that their scalar triple product must be zero. The scalar triple product of three vectors \( \vec{a} \), \( \vec{b} \), and \( \vec{c} \) is given by \( \vec{a} \cdot (\vec{b} \times \vec{c}) \). This value is equivalent to the determinant of the matrix formed by the components of the three vectors.

Setting up the Determinant for Coplanarity

We are given three vectors:

• \( \vec{v}_1 = {\rm{\alpha \hat i}} + {\rm{\alpha \hat j}} + {\rm{\gamma \hat k}} \)
• \( \vec{v}_2 = {\rm{\hat i}} + {\rm{\hat k}} = 1 \hat{i} + 0 \hat{j} + 1 \hat{k} \)
• \( \vec{v}_3 = {\rm{\gamma \hat i}} + {\rm{\gamma \hat j}} + {\rm{\beta \hat k}} \)

For these vectors to be coplanar, the determinant of the matrix formed by their components must be zero:

\( \begin{vmatrix} \alpha & \alpha & \gamma \\ 1 & 0 & 1 \\ \gamma & \gamma & \beta \end{vmatrix} = 0 \)

Calculating the Determinant and Solving for γ

Let's calculate the determinant. We can expand along the first row:

\( \alpha \left| \begin{matrix} 0 & 1 \\ \gamma & \beta \end{matrix} \right| - \alpha \left| \begin{matrix} 1 & 1 \\ \gamma & \beta \end{matrix} \right| + \gamma \left| \begin{matrix} 1 & 0 \\ \gamma & \gamma \end{matrix} \right| = 0 \)

Calculate the 2x2 determinants:

\( \alpha ((0)(\beta) - (1)(\gamma)) - \alpha ((1)(\beta) - (1)(\gamma)) + \gamma ((1)(\gamma) - (0)(\gamma)) = 0 \)

\( \alpha (0 - \gamma) - \alpha (\beta - \gamma) + \gamma (\gamma - 0) = 0 \)

\( -\alpha\gamma - \alpha\beta + \alpha\gamma + \gamma^2 = 0 \)

The terms \( -\alpha\gamma \) and \( +\alpha\gamma \) cancel each other out:

\( -\alpha\beta + \gamma^2 = 0 \)

Rearranging the equation to solve for \( \gamma \):

\( \gamma^2 = \alpha\beta \)

Since \( \alpha, \beta, \) and \( \gamma \) are given as non-negative numbers, we can take the square root of both sides:

\( \gamma = \sqrt{\alpha\beta} \)

Identifying the Relationship

The result \( \gamma = \sqrt{\alpha\beta} \) indicates a specific type of mean. For two non-negative numbers \( a \) and \( b \):

• Arithmetic Mean (AM) = \( \frac{a+b}{2} \)
• Geometric Mean (GM) = \( \sqrt{ab} \)
• Harmonic Mean (HM) = \( \frac{2}{\frac{1}{a} + \frac{1}{b}} = \frac{2ab}{a+b} \)

Our result \( \gamma = \sqrt{\alpha\beta} \) matches the definition of the geometric mean of \( \alpha \) and \( \beta \).

The question states that \( \alpha, \beta, \) and \( \gamma \) are distinct. If \( \alpha = \beta \), then \( \gamma = \sqrt{\alpha^2} = \alpha \), which would mean \( \alpha = \beta = \gamma \), contradicting the distinctness condition. Therefore, the distinctness condition implies that \( \alpha \neq \beta \), which naturally leads to \( \gamma \) being distinct from \( \alpha \) and \( \beta \) (unless one is zero and the other non-zero, or both are zero, but they are distinct non-negative numbers). The coplanarity condition yields the geometric mean relationship.

Conclusion

Based on the calculation using the coplanar vector condition, \( \gamma^2 = \alpha\beta \), which means \( \gamma \) is the geometric mean of \( \alpha \) and \( \beta \).

Revision Table: Key Concepts Revisited

Additional Information: Means and their Properties

For any two distinct positive numbers \( \alpha \) and \( \beta \), the Arithmetic Mean (AM), Geometric Mean (GM), and Harmonic Mean (HM) have a specific relationship:

• AM \( > \) GM \( > \) HM
• GM\(^2\) = AM \( \times \) HM (This property holds for two numbers)

The relationship \( \gamma = \sqrt{\alpha\beta} \) directly identifies \( \gamma \) as the geometric mean when \( \alpha \) and \( \beta \) are non-negative. The condition that \( \alpha, \beta, \gamma \) are distinct ensures that we are not in a trivial case where all are equal (which would happen if \( \alpha = \beta \), making AM=GM=HM). The coplanarity condition uniquely determines that \( \gamma \) must be the geometric mean of \( \alpha \) and \( \beta \) under the given distinct non-negative conditions.