MCQMediumJEE Main 2024 · 6 April, Shift 1Cross Product

Mathematics Question from JEE Main 2024 · 6 April, Shift 1

Let a⃗=2i^+αj^+k^\vec{a} = 2\hat{i} + \alpha \hat{j} + \hat{k}, b⃗=−i^+k^\vec{b} = -\hat{i} + \hat{k}, c⃗=βj^−k^\vec{c} = \beta \hat{j} - \hat{k}, where α\alpha and β\beta are integers and αβ=−6\alpha\beta = -6. Let the values of the ordered pair (α,β)(\alpha, \beta) for which the area of the parallelogram of diagonals a⃗+b⃗\vec{a} + \vec{b} and b⃗+c⃗\vec{b} + \vec{c} is 212\frac{\sqrt{21}}{2}, be (α1,β1)(\alpha_1, \beta_1) and (α2,β2)(\alpha_2, \beta_2). Then α12+β12−α2β2\alpha_1^2 + \beta_1^2 - \alpha_2\beta_2 is equal to:

  • A

    1717

  • B

    2424

  • C

    2121

  • D

    1919

Answer & step-by-step solution

Sign in to reveal the correct answer, the full step-by-step solution, and the common mistakes for this question.

Practice more Cross Product questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step - free to start.

Related questions