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Area of a quadrilateral or parallelogram as half the cross product of its diagonals

This type of question, asked 6 times: 1 Feb 2023, 8 Apr 2023, 30 Jan 2024, 5 Apr 2024, 6 Apr 2024, 9 Apr 2024. All 6 questions below, oldest first.

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  1. Question 1

    1 Feb 2023

    90
    A(2,6,2),B(−4,0,λ),C(2,3,−1)A(2,6,2),B( - 4,0,\lambda),C(2,3, - 1) and D(4,5,0),∣λ∣≤5D(4,5,0),|\lambda| \leq 5 are the vertices of a quadrilateral ABCDABCD. If its area is 18 square units, then 5−6λ5 - 6\lambda is equal to

  2. Question 2

    8 Apr 2023

    161
    The area of the quadrilateral ABCDABCD with vertices A(2,1,1),B(1,2,5),C(−2,−3,5)A(2,1,1),B(1,2,5),C( - 2, - 3,5) and D(1,−6,−D(1, - 6, - 7) is equal to

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  3. Question 3

    30 Jan 2024

    78
    Let A(2,3,5)A(2,3,5) and C(−3,4,−2)C( - 3,4, - 2) be opposite vertices of a parallelogram ABCDABCD if the diagonal BD→=i^+2j^+3k^\overrightarrow{BD}=\widehat{i}+ 2\widehat{j}+ 3\widehat{k} then the area of the parallelogram is equal to

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  4. Question 4

    5 Apr 2024

    78
    If A(1,−1,2),B(5,7,−6),C(3,4,−10)A(1, - 1,2),B(5,7, - 6),C(3,4, - 10) and D(−1,−4,−2)D( - 1, - 4, - 2) are the vertices of a quadrilateral ABCDABCD, then its area is :

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  5. Question 5

    6 Apr 2024

    62
    If A(3,1,−1),B(53,73,13),C(2,2,1)A(3,1, - 1),B\left( \frac{5}{3},\frac{7}{3},\frac{1}{3} \right),C(2,2,1) and D(103,23,−13)D\left( \frac{10}{3},\frac{2}{3},\frac{- 1}{3} \right) are the vertices of a quadrilateral ABCDABCD, then its area is

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  6. Question 6

    9 Apr 2024

    170
    Let a→=2i^+αj^+k^, b→=−i^+k^, c→=βj^−k^\overrightarrow{a}= 2\widehat{i}+ \alpha\widehat{j}+\widehat{k},\ \overrightarrow{b}= -\widehat{i}+\widehat{k},\ \overrightarrow{c}= \beta\widehat{j}-\widehat{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 a→+b→\overrightarrow{a}+\overrightarrow{b} and b→+c→\overrightarrow{b}+\overrightarrow{c} is 212\frac{\sqrt{21}}{2}, be (α1,β1)\left( \alpha_{1},\beta_{1} \right) and (α2,β2)\left( \alpha_{2},\beta_{2} \right). Then α12+β12−α2β2\alpha_{1}^{2}+\beta_{1}^{2}-\alpha_{2}\beta_{2} is equal to

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