MCQMediumJEE Main 2023 · 6 April, Shift 1Cross Product

Mathematics Question from JEE Main 2023 · 6 April, Shift 1

Let a=2i^+3j^+4k^\vec{a} = 2\hat{i} + 3\hat{j} + 4\hat{k}, b=i^2j^2k^\vec{b} = \hat{i} - 2\hat{j} - 2\hat{k} and c=i^+4j^+3k^\vec{c} = -\hat{i} + 4\hat{j} + 3\hat{k}. If d\vec{d} is a vector perpendicular to both b\vec{b} and c\vec{c}, and ad=18\vec{a} \cdot \vec{d} = 18, then a×d2|\vec{a} \times \vec{d}|^2 is equal to :

  • A

    760760

  • B

    640640

  • C

    720720

  • D

    680680

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