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:
- 17
- 24
- 21
- 19
The Correct Option is D
Approach Solution - 1
To solve the given problem, let's first express the diagonals of the parallelogram. We have:
\(\vec{a} = 2\hat{i} + \alpha \hat{j} + \hat{k}\)
\(\vec{b} = -\hat{i} + \hat{k}\)
\(\vec{c} = \beta \hat{j} - \hat{k}\)
The diagonals are given by:
\(\vec{d}_1 = \vec{a} + \vec{b} = (2-1)\hat{i} + \alpha \hat{j} + (1+1)\hat{k} = \hat{i} + \alpha \hat{j} + 2\hat{k}\)
\(\vec{d}_2 = \vec{b} + \vec{c} = (-1 + 0)\hat{i} + \beta \hat{j} + (1-1)\hat{k} = -\hat{i} + \beta \hat{j}\)
The area of the parallelogram from the diagonals is given by half the magnitude of the cross product of the two diagonal vectors:
\(\text{Area} = \frac{1}{2} |\vec{d}_1 \times \vec{d}_2|\)
Calculate the cross product:
\(\vec{d}_1 \times \vec{d}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & \alpha & 2 \\ -1 & \beta & 0 \end{vmatrix}\)
Expanding this determinant, we have:
\(= \hat{i}(0 - 2\beta) - \hat{j}(2 - 0) + \hat{k}(1\beta - (-\alpha))\)
\(= -2\beta \hat{i} - 2\hat{j} + (\beta + \alpha)\hat{k}\)
The magnitude of this vector is:
\(|\vec{d}_1 \times \vec{d}_2| = \sqrt{(-2\beta)^2 + (-2)^2 + (\beta + \alpha)^2}\)
\(= \sqrt{4\beta^2 + 4 + \beta^2 + 2\alpha\beta + \alpha^2}\)
\(= \sqrt{5\beta^2 + 2\alpha\beta + \alpha^2 + 4}\)
The area is given as:
\(\frac{\sqrt{21}}{2}\)
Therefore, we equate:
\(\frac{1}{2}\sqrt{5\beta^2 + 2\alpha\beta + \alpha^2 + 4} = \frac{\sqrt{21}}{2}\)
\(\sqrt{5\beta^2 + 2\alpha\beta + \alpha^2 + 4} = \sqrt{21}\)
On squaring both sides, we get:
\(5\beta^2 + 2\alpha\beta + \alpha^2 + 4 = 21\)
\(5\beta^2 + 2\alpha\beta + \alpha^2 = 17\)
Given \(\alpha \beta = -6\), let's substitute:
Possible integer pairs that satisfy \(\alpha \beta = -6\) are: \((\alpha, \beta) = (2, -3), (-2, 3), (3, -2), (-3, 2), (6, -1), (-6, 1), (1, -6), (-1, 6)\)
Substitute these pairs into \(5\beta^2 + 2\alpha\beta + \alpha^2 = 17\) to find correct such pair.
Checking the pair \((\alpha, \beta) = (3, -2)\):
\(5(-2)^2 + 2(3)(-2) + 3^2 = 5(4) - 12 + 9 = 20 - 12 + 9 = 17\)
Check another pair \((\alpha, \beta) = (-3, 2)\):
\(5(2)^2 + 2(-3)(2) + (-3)^2 = 5(4) - 12 + 9 = 17\)
These two pairs satisfy the condition: \((\alpha_1, \beta_1) = (3, -2)\) and \((\alpha_2, \beta_2) = (-3, 2)\)
Now, compute \(\alpha_1^2 + \beta_1^2 - \alpha_2\beta_2\)\):
\((3)^2 + (-2)^2 - (-3)(2) = 9 + 4 + 6 = 19\)
Thus, the answer is:
Option: 19
Approach Solution -2
We are given the following vectors:
\(\mathbf{a} = 2\mathbf{i} + \alpha \mathbf{j} + \mathbf{k}, \quad \mathbf{b} = -\mathbf{i} + \mathbf{j} + \mathbf{k}, \quad \mathbf{c} = \beta \mathbf{j} - \mathbf{k}\)
We are also given that \(\alpha \beta = -6\) and need to find \(\alpha_1^2 + \beta_2^2 - \alpha_2 \beta_2\), where the area of the parallelogram formed by the diagonals \(\mathbf{a} + \mathbf{b}\) and \(\mathbf{b} + \mathbf{c}\) is \(\frac{\sqrt{21}}{2}\).
Step 1: Area of the Parallelogram
The area of a parallelogram formed by vectors \(\mathbf{u}\) and \(\mathbf{v}\) is given by the magnitude of the cross product \(|\mathbf{u} \times \mathbf{v}|\). In this case, the vectors are \(\mathbf{u} = \mathbf{a} + \mathbf{b}\) and \(\mathbf{v} = \mathbf{b} + \mathbf{c}\). Thus, we need to compute the cross product \((\mathbf{a} + \mathbf{b}) \times (\mathbf{b} + \mathbf{c})\).
Step 2: Compute the Cross Product
First, calculate \(\mathbf{a} + \mathbf{b}\) and \(\mathbf{b} + \mathbf{c}\):
\(\mathbf{a} + \mathbf{b} = (2 - 1)\mathbf{i} + (\alpha + 1)\mathbf{j} + (1 + 1)\mathbf{k} = \mathbf{i} + (\alpha + 1)\mathbf{j} + 2\mathbf{k}\)
\(\mathbf{b} + \mathbf{c} = (-1 + 0)\mathbf{i} + (1 + \beta)\mathbf{j} + (1 - 1)\mathbf{k} = -\mathbf{i} + (1 + \beta)\mathbf{j}\)
Now, calculate the cross product \((\mathbf{a} + \mathbf{b}) \times (\mathbf{b} + \mathbf{c})\):
\[ \mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & \alpha + 1 & 2 \\ -1 & 1 + \beta & 0 \end{vmatrix} \]
Expanding the determinant, we get:
\[ \mathbf{u} \times \mathbf{v} = \mathbf{i} \left( (\alpha + 1)(0) - 2(1 + \beta) \right) - \mathbf{j} \left( 1(0) - 2(-1) \right) + \mathbf{k} \left( 1(1 + \beta) - (-1)(\alpha + 1) \right) \]
After calculating each determinant, we get:
\[ \mathbf{u} \times \mathbf{v} = \mathbf{i}[-2(1 + \beta)] - \mathbf{j}[-2] + \mathbf{k}[1 + \beta + \alpha + 1] \]
Simplifying:
\[ \mathbf{u} \times \mathbf{v} = (-2\beta)\mathbf{i} + 2\mathbf{j} + (1 + \beta + \alpha)\mathbf{k} \]
Step 3: Set the Area Equal to \(\frac{\sqrt{21}}{2}\)
The magnitude of this vector is:
\[ |\mathbf{u} \times \mathbf{v}| = \sqrt{(-2\beta)^2 + 2^2 + (1 + \beta + \alpha)^2} \]
Solving this equation will yield the values for \(\alpha\) and \(\beta\).
Step 4: Use the Relationship \(\alpha \beta = -6\)
Using the relationship \(\alpha \beta = -6\), we solve the system of equations to find the values of \(\alpha\) and \(\beta\).
The possible pairs are \((\alpha_1, \beta_1) = (3, -2)\) and \((\alpha_2, \beta_2) = (-2, 3)\).
Step 5: Final Calculation
Now, we calculate \(\alpha_1^2 + \beta_2^2 - \alpha_2 \beta_2\):
\[ \alpha_1^2 + \beta_2^2 - \alpha_2 \beta_2 = 3^2 + (-2)^2 - (-2)(3) = 9 + 4 + 6 = 19 \]
Thus, the correct answer is: 19
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