JEE Mathematics 2024 Question with Solution

Given:

$$\vec{a} = 2\hat{i} + \alpha \hat{j} + \hat{k}, \quad \vec{b} = -\hat{i} + \hat{k}, \quad \vec{c} = \beta \hat{j} - \hat{k}, \quad \alpha\beta = -6$$

The diagonals are $\vec{a}+\vec{b}$ and $\vec{b}+\vec{c}$, and the area of the parallelogram is $\frac{\sqrt{21}}{2}$.

Find: The value of $\alpha_1^2 + \beta_1^2 - \alpha_2\beta_2$.

First compute the two diagonal vectors:

$$\vec{d}_1 = \vec{a} + \vec{b} = (2-1)\hat{i} + \alpha\hat{j} + (1+1)\hat{k} = \hat{i} + \alpha\hat{j} + 2\hat{k}$$ $$\vec{d}_2 = \vec{b} + \vec{c} = -\hat{i} + \beta\hat{j}$$

For a parallelogram whose diagonals are $\vec{d}_1$ and $\vec{d}_2$, area is:

$$\text{Area} = \frac{1}{2}\left|\vec{d}_1 \times \vec{d}_2\right|$$

Now evaluate the cross product:

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

Therefore,

$$\left|\vec{d}_1 \times \vec{d}_2\right| = \sqrt{(-2\beta)^2 + (-2)^2 + (\alpha+\beta)^2}$$ $$= \sqrt{4\beta^2 + 4 + \alpha^2 + 2\alpha\beta + \beta^2}$$ $$= \sqrt{5\beta^2 + 2\alpha\beta + \alpha^2 + 4}$$

Using the given area,

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

So,

$$5\beta^2 + 2\alpha\beta + \alpha^2 + 4 = 21$$ $$5\beta^2 + 2\alpha\beta + \alpha^2 = 17$$

Using $\alpha\beta=-6$,

$$5\beta^2 + \alpha^2 - 12 = 17$$ $$5\beta^2 + \alpha^2 = 29$$

Now test integer pairs satisfying $\alpha\beta=-6$. The pairs $(3,-2)$ and $(-3,2)$ satisfy the equation:

$$5(-2)^2 + 3^2 = 20 + 9 = 29$$ $$5(2)^2 + (-3)^2 = 20 + 9 = 29$$

Hence,

$$(\alpha_1,\beta_1) = (3,-2), \quad (\alpha_2,\beta_2) = (-3,2)$$

Now compute:

$$\alpha_1^2 + \beta_1^2 - \alpha_2\beta_2 = 3^2 + (-2)^2 - (-3)(2)$$ $$= 9 + 4 + 6 = 19$$

Therefore, the correct option is D.

Since $\alpha\beta=-6$, the possible integer pairs are:

$$(2,-3),\ (-2,3),\ (3,-2),\ (-3,2),\ (6,-1),\ (-6,1),\ (1,-6),\ (-1,6)$$

From the derived condition,

$$5\beta^2 + 2\alpha\beta + \alpha^2 = 17$$

Substitute pairwise:

$$(3,-2): \quad 5(4) + 2(3)(-2) + 9 = 20 - 12 + 9 = 17$$ $$(-3,2): \quad 5(4) + 2(-3)(2) + 9 = 20 - 12 + 9 = 17$$

These are the two required ordered pairs. Then,

$$\alpha_1^2 + \beta_1^2 - \alpha_2\beta_2 = 9 + 4 + 6 = 19$$

So the correct option is D.

Note: The source question asks for $\alpha_1 + \beta_1 - \alpha_2\beta_2$, but the solution consistently computes $\alpha_1^2 + \beta_1^2 - \alpha_2\beta_2$ and matches option $19$. The answer has therefore been derived from the solution, which is the primary source.