Let , , , where and are integers and . Let the values of the ordered pair , for which the area of the parallelogram of diagonals and is , be and . Then is equal to [2024]

Question

Let $\vec{a} = 2 \hat{i} + α \hat{j} + \hat{k}$ , $\vec{b} = - \hat{i} + \hat{k}$ , $\vec{c} = β \hat{j} - \hat{k}$ , where $α$ and $β$ are integers and $α β = - 6$ . Let the values of the ordered pair $(α, β)$ , for which the area of the parallelogram of diagonals $\vec{a} + \vec{b}$ and $\vec{b} + \vec{c}$ is $\frac{\sqrt{21}}{2}$ , be $(α_{1}, β_{1})$ and $(α_{2}, β_{2})$ . Then $α_{1}^{2} + β_{1}^{2} - α_{2} β_{2}$ is equal to [2024]

Smart Achievers · Accepted Answer

(4)

Area of parallelogram $= \frac{1}{2} \times$ $| (\vec{a} + \vec{b}) \times (\vec{b} + \vec{c}) |$

$\vec{c} = β \hat{j} - \hat{k}$ $| (\vec{a} + \vec{b}) \times (\vec{b} + \vec{c}) |$

$\Rightarrow \sqrt{21} =$ $| (\hat{i} + α \hat{j} + 2 \hat{k}) \times (- \hat{i} + β \hat{j}) |$

$\Rightarrow \sqrt{21} =$ $| | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & α & 2 \\ - 1 & β & 0 \end{matrix} | |$

$\Rightarrow \sqrt{21} =$ $| - 2 β \hat{i} - 2 \hat{j} + (α + β) \hat{k} |$

$\Rightarrow α^{2} + 5 β^{2} + 2 α β = 17$

$\Rightarrow α^{2} + 5 β^{2} = 17 + 12$ $[∵ α β = - 6]$

$\Rightarrow α^{2} + 5 β^{2} = 29$

and $α β = - 6, α, β$ are integers.

$\Rightarrow α = - 3, β = 2 or α = 3, β = - 2$

$α_{1}^{2} + β_{1}^{2} - α_{2} β_{2} = 9 + 4 + 6 = 19$ .

Solution

1 21

2 17

3 24

4 19

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