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Use |a x b|^2 + (a . b)^2 = |a|^2 |b|^2

This type of question, asked 5 times: 2021 Paper 15, 25 Jun 2022, 29 Jul 2022, 31 Jan 2023, 30 Jan 2024. All 5 questions below, oldest first.

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  1. Question 1

    2021 Paper 15

    49
    If ∣a→∣=2,∣b→∣=5,∣a→×b→∣=8|\overrightarrow{a}| = 2,|\overrightarrow{b}| = 5,|\overrightarrow{a}\times\overrightarrow{b}| = 8, then ∣a→⋅b→∣=|\overrightarrow{a}\cdot\overrightarrow{b}| = ?

  2. Question 2

    25 Jun 2022

    83
    Let θ\theta be the angle between the vectors a→\overrightarrow{a} and b→\overrightarrow{b}, where ∣a→∣=4,∣b→∣=3 θ∈(π4,π3)|\overrightarrow{a}| = 4,|\overrightarrow{b}| = 3\ \theta \in\left( \frac{\pi}{4},\frac{\pi}{3} \right). Then ∣(a→−b→)×(a→+b→)∣2+4(a→⋅b→)2|(\overrightarrow{a}-\overrightarrow{b}) \times (\overrightarrow{a}+\overrightarrow{b})|^{2}+ 4(\overrightarrow{a}\cdot\overrightarrow{b})^{2} is equal to

  3. Question 3

    29 Jul 2022

    179
    Let a→\overrightarrow{a} and b→\overrightarrow{b} be two vectors such that ∣a→+b→∣2=∣a→∣2+2∣b→∣2,a→⋅b→=3|\overrightarrow{a}+\overrightarrow{b}|^{2}= |\overrightarrow{a}|^{2}+ 2|\overrightarrow{b}|^{2},\overrightarrow{a}\cdot\overrightarrow{b}= 3 and ∣a→×b→∣2=75|\overrightarrow{a}\times\overrightarrow{b}|^{2}= 75. Then ∣a→∣2|\overrightarrow{a}|^{2} is equal to

  4. Question 4

    31 Jan 2023

    87
    Let a→\overrightarrow{a} and b→\overrightarrow{b} be two vectors such that ∣a→∣=14,∣b→∣=6|\overrightarrow{a}| =\sqrt{14},|\overrightarrow{b}| =\sqrt{6} and ∣a→×b→∣=48|\overrightarrow{a}\times\overrightarrow{b}| =\sqrt{48}. Then (a→⋅b→)2(\overrightarrow{a}\cdot\overrightarrow{b})^{2} is equal to

  5. Question 5

    30 Jan 2024

    154
    Let a→=i^+αj^+βk^,α,β∈R\overrightarrow{a}=\widehat{i}+\alpha\widehat{j}+\beta\widehat{k},\alpha,\beta\in R. Let a vector b→\overrightarrow{b} be such that the angle between a→\overrightarrow{a} and b→\overrightarrow{b} is π4\frac{\pi}{4} and ∣b→∣2=6|\overrightarrow{b}|^{2}= 6, If a→⋅b→=32\overrightarrow{a}\cdot\overrightarrow{b}= 3\sqrt{2}, then the value of (α2+β2)∣a→×b→∣2\left( \alpha^{2}+\beta^{2} \right)|\overrightarrow{a}\times\overrightarrow{b}|^{2} is equal to

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