Let \overrightarrow{a}=2 \stackrel{\^}{i}+\alpha \stackrel{\^}{j}+\stackrel{\^}{k}, \overrightarrow{b}=-\stackrel{\^}{i}+\stackrel{\^}{k}, \overrightarrow{c}=\beta \stackrel{\^}{j}-\stackrel{\^}{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 \overrightarrow{a}+\overrightarrow{b} and \overrightarrow{b}+\overrightarrow{c} is {\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 [9 Apr 2024 Shift 2]

JEE Main 9 Apr 2024 Shift 2 Solved Paper

Section: Mathematics

Question : 20 of 90

Marks: +1, -0

Let

→

+α

,‌‌

→

=−

,‌‌

→

=β

−

, where α and β are integers and αβ=−6. Let the values of the ordered pair (α,β) for which the area of the parallelogram of diagonals

→

and

→

is ‌

√21

, be (α1,β1) and (α2,β2).
Then α12+β12−α2β2 is equal to

[9 Apr 2024 Shift 2]

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