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a + b + c = 0 with given magnitudes: angles between the vectors

This type of question, asked 4 times: 14 May 2024 S1, 15 Apr 2026 S1, 15 Apr 2026 S2, 15 May 2026 S1. All 4 questions below, oldest first.

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  1. Question 1

    14 May 2024 S1

    115
    If a⃗,b⃗,c⃗\vec{a},\vec{b},\vec{c} are unit vectors and θ\theta is angle between a⃗\vec{a} and c⃗\vec{c} and a⃗+2b⃗+2c⃗=0⃗\vec{a}+2\vec{b}+2\vec{c}=\vec{0}, then ∣a⃗×c⃗∣=|\vec{a}\times\vec{c}|=

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  2. Question 2

    15 Apr 2026 S1

    91
    Let (pˉ∧qˉ)(\bar{p} \wedge \bar{q}) denote the angle between pˉ\bar{p} and qˉ\bar{q}. If aˉ+bˉ+cˉ=0ˉ,∣aˉ∣=7,∣bˉ∣=5\bar{a} + \bar{b} + \bar{c} = \bar{0}, |\bar{a}| = 7, |\bar{b}| = 5 and ∣cˉ∣=3|\bar{c}| = 3 then (take π=227\pi = \dfrac{22}{7})

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  3. Question 3

    15 Apr 2026 S2

    88
    If a⃗+b⃗+c⃗=0⃗,∣a⃗∣=∣b⃗∣=∣c⃗∣=3\vec{a} + \vec{b} + \vec{c} = \vec{0}, |\vec{a}| = |\vec{b}| = |\vec{c}| = 3 and θ\theta is the angle between b⃗\vec{b} and c⃗\vec{c} then tan⁡2θ+cot⁡2θ=\tan^2\theta + \cot^2\theta =

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  4. Question 4

    15 May 2026 S1

    90
    If ∣aˉ∣=24|\bar{a}| = 24, ∣bˉ∣=7|\bar{b}| = 7, aˉ⋅bˉ=0\bar{a} \cdot \bar{b} = 0, aˉ+bˉ+cˉ=0ˉ\bar{a} + \bar{b} + \bar{c} = \bar{0}, then ∣cˉ∣=…|\bar{c}| = \ldots

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