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

This type of question, asked 6 times: 3 May 2023 S1, 25 Apr 2025 S2, 17 Apr 2026 S2, 18 Apr 2026 S2, 12 May 2026 S2, 14 May 2026 S1. All 6 questions below, oldest first.

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  1. Question 1

    3 May 2023 S1

    114
    If ∣a⃗∣=27|\vec{a}|=\sqrt{27}, ∣b⃗∣=7|\vec{b}|=7 and ∣a⃗×b⃗∣=35|\vec{a}\times\vec{b}|=35, then a⃗⋅b⃗\vec{a}\cdot\vec{b} is equal to

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  2. Question 2

    25 Apr 2025 S2

    109
    If θ\theta is an obtuse angle between vectors aˉ\bar{a} and b‾\overline{b} such that ∣a‾∣=5,∣ b‾∣=3|\overline{a}| = 5,|\overline{\text{ }b}| = 3 and ∣a‾×b‾∣=55|\overline{a}\times\overline{b}| = 5\sqrt{5} then aˉ⋅bˉ=\bar{a}\cdot\bar{b}=

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  3. Question 3

    17 Apr 2026 S2

    90
    If ∣a⃗⋅b⃗∣=∣a⃗×b⃗∣,a⃗⋅b⃗<0|\vec{a}\cdot\vec{b}| = |\vec{a} \times \vec{b}|, \vec{a}\cdot\vec{b} < 0 and θ\theta is the angle between a⃗\vec{a} and b⃗\vec{b}, then the value of sin⁡θ+tan⁡θ\sin\theta + \tan\theta is...

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  4. Question 4

    18 Apr 2026 S2

    92
    The value of ∣aˉ⋅bˉ∣2+∣aˉ×bˉ∣⋅∣aˉ×bˉ∣|\bar{a}\cdot\bar{b}|^2 + |\bar{a} \times \bar{b}| \cdot |\bar{a} \times \bar{b}| is …\ldots

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  5. Question 5

    12 May 2026 S2

    91
    If ∣u⃗∣=5,∣v⃗∣=13,∣u⃗×v⃗∣=25|\vec{u}| = 5, |\vec{v}| = 13, |\vec{u} \times \vec{v}| = 25, then u⃗⋅v⃗=…\vec{u}\cdot\vec{v} = \ldots

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  6. Question 6

    14 May 2026 S1

    89
    For two non-zero, non-collinear, non-orthogonal vectors a⃗\vec{a} and b⃗\vec{b} where θ\theta is angle between them, then (a⃗⋅b⃗)2+∣a⃗×b⃗∣2(a⃗⋅b⃗)2−∣a⃗×b⃗∣2=\dfrac{(\vec{a}\cdot\vec{b})^2 + |\vec{a} \times \vec{b}|^2}{(\vec{a}\cdot\vec{b})^2 - |\vec{a} \times \vec{b}|^2} =

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