Let a=2i+ j+k,b=-i+k,c= j-k, where and are integers and =-6.

Question

Let $\vec{a}=2\hat{i}+\alpha \hat{j}+\hat{k},\vec{b}=-\hat{i}+\hat{k},\vec{c}=\beta \hat{j}-\hat{k}$, 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 ${\alpha }_{2}^{1}+{\beta }_{2}^{1}-{\alpha }_{2}{\beta }_{2}$ is equal to

Accepted Answer

Area of parallelogram whose diagonals are $\vec{a}+\vec{b}$ and $\vec{b}+\vec{c}$ is

$\begin{gathered} =\frac{1}{2}|(\vec{a}+\vec{b})	imes (\vec{b}+\vec{c})| \ =\frac{1}{2}|\vec{a}	imes \vec{b}+\vec{a}	imes \vec{c}+\vec{b}	imes \vec{c}| \ =\frac{1}{2}|-2\beta \hat{i}-2\hat{j}+(\alpha +\beta )\hat{k}| \end{gathered}$

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

Which is given $\frac{\sqrt{21}}{2}$

$\begin{gathered} 	herefore 4{\beta }^{2}+4+(\alpha +\beta {)}^{2}=21 \ \Rightarrow (\alpha +\beta {)}^{2}+4{\beta }^{2}=17 \ \Rightarrow {\alpha }^{2}+5{\beta }^{2}+2\alpha \beta =17 \ \Rightarrow {\alpha }^{2}+5{\beta }^{2}=29 \ 	herefore (\alpha ,\beta )\in {(3,2),(-3,-2),(-3,2),(3,-2)} \ \because \alpha \beta =-6 \ 	herefore (\alpha ,\beta )\in {(-3,2),(3,-2)} \ 	herefore {\alpha }_{2}^{1}+{\beta }_{2}^{1}-{\alpha }_{2}{\beta }_{2} \ = 9+4-(-6)=19 \end{gathered}$

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