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a = (3i + k)/sqrt(10), b = (2i + 3j - 6k)/7: value of (2a - b).[(a x b) x (a + 2b)]

The same question, asked 4 times: 3 May 2023 S2*, 11 May 2023 S2, 13 May 2024 S2, 20 Apr 2025 S2. Every wording below, oldest first.

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  1. Question 1

    3 May 2023 S2

    108
    If a⃗=110(4i^−3j^+k^)\vec{a}=\frac{1}{\sqrt{10}}(4\hat{i}-3\hat{j}+\hat{k}), b⃗=15(i^+2j^+2k^)\vec{b}=\frac{1}{5}(\hat{i}+2\hat{j}+2\hat{k}), then the value of (2a⃗−b⃗)⋅{(a⃗×b⃗)×(a⃗+2b⃗)}(2\vec{a}-\vec{b})\cdot\{(\vec{a}\times\vec{b})\times(\vec{a}+2\vec{b})\} is

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  2. Question 2

    11 May 2023 S2

    113
    If a⃗=110(3i^+k^)\vec{a}=\frac{1}{\sqrt{10}}(3\hat{i}+\hat{k}) and b⃗=17(2i^+3j^−6k^)\vec{b}=\frac{1}{7}(2\hat{i}+3\hat{j}-6\hat{k}), then the value of (2a⃗−b⃗)⋅[(a⃗×b⃗)×(a⃗+2b⃗)](2\vec{a}-\vec{b})\cdot[(\vec{a}\times\vec{b})\times(\vec{a}+2\vec{b})] is

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  3. Question 3

    13 May 2024 S2

    142
    If a⃗=110(3i^+k^),b⃗=17(2i^+3j^−6k^)\vec{a}=\frac{1}{\sqrt{10}}(3\hat{i}+\hat{k}), \vec{b}=\frac{1}{7}(2\hat{i}+3\hat{j}-6\hat{k}), then the value of (2a⃗−b⃗)⋅[(a⃗×b⃗)×(a⃗+2b⃗)](2\vec{a}-\vec{b})\cdot[(\vec{a}\times\vec{b})\times(\vec{a}+2\vec{b})] is:

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  4. Question 4

    20 Apr 2025 S2

    147
    If aˉ=110(3i^+k^)\bar{a}=\frac{1}{\sqrt{10}}(3\widehat{i}+\widehat{k}) and bˉ=17(2i^+3j^−6k^)\bar{b}=\frac{1}{7}(2\widehat{i}+ 3\widehat{j}- 6\widehat{k}) then the value of (2aˉ−bˉ)⋅[(aˉ×bˉ)×(aˉ+2bˉ)]=(2\bar{a}-\bar{b}) \cdot \lbrack(\bar{a}\times\bar{b}) \times (\bar{a}+ 2\bar{b})\rbrack =

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