Question

Let \[ \vec{a} = 2\hat{i} + \alpha \hat{j} + \hat{k}, \quad \vec{b} = -\hat{i} + \hat{k}, \quad \vec{c} = \beta \hat{j} - \hat{k}, \] where \( \alpha \) and \( \beta \) are integers and \( \alpha \beta = -6 \). Let the values of the ordered pair \( (\alpha, \beta) \) for which the area of the parallelogram of diagonals \( \vec{a} + \vec{b} \) and \( \vec{b} + \vec{c} \) is \( \frac{\sqrt{21}}{2} \), be \( (\alpha_1, \beta_1) \) and \( (\alpha_2, \beta_2) \). Then \( \alpha_1^2 + \beta_1^2 - \alpha_2 \beta_2 \) is equal to:

• A. 17
• B. 24
• C. 21
• D. 19

Answer 1

The Correct Option is D

To resolve the stated problem, the diagonals of the parallelogram are first defined. The vectors representing the sides are given as:

\(\vec{a} = 2\hat{i} + \alpha \hat{j} + \hat{k}\)

\(\vec{b} = -\hat{i} + \hat{k}\)

\(\vec{c} = \beta \hat{j} - \hat{k}\)

The diagonals (\(\vec{d}_1\) and \(\vec{d}_2\)) are calculated as:

\(\vec{d}_1 = \vec{a} + \vec{b} = (2-1)\hat{i} + \alpha \hat{j} + (1+1)\hat{k} = \hat{i} + \alpha \hat{j} + 2\hat{k}\)

\(\vec{d}_2 = \vec{b} + \vec{c} = (-1 + 0)\hat{i} + \beta \hat{j} + (1-1)\hat{k} = -\hat{i} + \beta \hat{j}\)

The area of the parallelogram formed by these diagonals is half the magnitude of their cross product:

\(\text{Area} = \frac{1}{2} |\vec{d}_1 \times \vec{d}_2|\)

The cross product is computed as follows:

\(\vec{d}_1 \times \vec{d}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & \alpha & 2 \\ -1 & \beta & 0 \end{vmatrix}\)

Expanding the determinant yields:

\(= \hat{i}(0 - 2\beta) - \hat{j}(2 - 0) + \hat{k}(1\beta - (-\alpha))\)

\(= -2\beta \hat{i} - 2\hat{j} + (\beta + \alpha)\hat{k}\)

The magnitude of this resulting vector is:

\(|\vec{d}_1 \times \vec{d}_2| = \sqrt{(-2\beta)^2 + (-2)^2 + (\beta + \alpha)^2}\)

\(= \sqrt{4\beta^2 + 4 + \beta^2 + 2\alpha\beta + \alpha^2}\)

\(= \sqrt{5\beta^2 + 2\alpha\beta + \alpha^2 + 4}\)

Given that the area is \(\frac{\sqrt{21}}{2}\), we set up the equation:

\(\frac{1}{2}\sqrt{5\beta^2 + 2\alpha\beta + \alpha^2 + 4} = \frac{\sqrt{21}}{2}\)

Simplifying this equation leads to:

\(\sqrt{5\beta^2 + 2\alpha\beta + \alpha^2 + 4} = \sqrt{21}\)

Squaring both sides gives:

\(5\beta^2 + 2\alpha\beta + \alpha^2 + 4 = 21\)

Which further simplifies to:

\(5\beta^2 + 2\alpha\beta + \alpha^2 = 17\)

Using the provided condition \(\alpha \beta = -6\), we substitute this value into the equation. The possible integer pairs for \((\alpha, \beta)\) that satisfy \(\alpha \beta = -6\) are: \((2, -3), (-2, 3), (3, -2), (-3, 2), (6, -1), (-6, 1), (1, -6), (-1, 6)\). These pairs are then substituted into \(5\beta^2 + 2\alpha\beta + \alpha^2 = 17\) to identify the correct pair(s).

Testing the pair \((\alpha, \beta) = (3, -2)\):

\(5(-2)^2 + 2(3)(-2) + 3^2 = 5(4) - 12 + 9 = 20 - 12 + 9 = 17\)

Testing the pair \((\alpha, \beta) = (-3, 2)\):

\(5(2)^2 + 2(-3)(2) + (-3)^2 = 5(4) - 12 + 9 = 17\)

Both pairs satisfy the condition. Thus, the valid pairs are \((\alpha_1, \beta_1) = (3, -2)\) and \((\alpha_2, \beta_2) = (-3, 2)\).

Finally, we compute \(\alpha_1^2 + \beta_1^2 - \alpha_2\beta_2\):

\((3)^2 + (-2)^2 - (-3)(2) = 9 + 4 + 6 = 19\)

The final answer is 19.

Option: 19