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Expand vector triple products with a x (b x c) = (a . c)b - (a . b)c

This type of question, asked 10 times: 2021 Paper 17, 2021 Paper 23, 2021 Paper 25, 27 Jul 2022, 29 Jul 2022, 25 Jan 2023, 30 Jan 2023, 1 Feb 2024, 6 Apr 2024, 23 Jan 2026 S1. All 11 questions below, oldest first.

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  1. Question 1

    2021 Paper 17

    78
    Let a→,b→\overrightarrow{a},\overrightarrow{b} and c→\overrightarrow{c} be three vectors such that a→=b→×(b→×c→)\overrightarrow{a}=\overrightarrow{b}\times (\overrightarrow{b}\times\overrightarrow{c}). If magnitudes of the vectors a→,b→\overrightarrow{a},\overrightarrow{b} and c→\overrightarrow{c} are 2,1\sqrt{2},1 and 2 respectively and the angle between b→\overrightarrow{b} and c→\overrightarrow{c} is θ(0<θ<π2)\theta\left( 0 <\theta<\frac{\pi}{2} \right), then the value of 1+tan⁡θ1 + \tan\theta is equal to :

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  2. Question 2

    2021 Paper 23

    62
    Let a→,b→,c→\overrightarrow{a},\overrightarrow{b},\overrightarrow{c} be three vectors mutually perpendicular to each other and have same magnitude. If a vector r→\overrightarrow{r} satisfies a→×{(r→−b→)×a→}+b→×{(r→−c→)×b→}+c→×{(r→−a→)×c→}=0→\overrightarrow{a}\times\{(\overrightarrow{r}-\overrightarrow{b})\times\overrightarrow{a}\}+\overrightarrow{b}\times\{(\overrightarrow{r}-\overrightarrow{c})\times\overrightarrow{b}\}+\overrightarrow{c}\times\{(\overrightarrow{r}-\overrightarrow{a})\times\overrightarrow{c}\}=\overrightarrow{0}, then r→\overrightarrow{r} is equal to:

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  3. Question 3

    2021 Paper 25

    65
    If a→&b→\overrightarrow{a}\&\overrightarrow{b} are perpendicular vectors, then a→×(a→×(a→×(a→×b→)))\overrightarrow{a}\times (\overrightarrow{a}\times (\overrightarrow{a}\times (\overrightarrow{a}\times\overrightarrow{b}))) is equal to

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  4. Question 4

    27 Jul 2022

    73
    Let a→=2i^−j^+5k^\overrightarrow{a}= 2\widehat{i}-\widehat{j}+ 5\widehat{k} and b→=αi^+βj^+2k^\overrightarrow{b}= \alpha\widehat{i}+ \beta\widehat{j}+ 2\widehat{k}. If ((a→×b→)×i^)⋅k^=232((\overrightarrow{a}\times\overrightarrow{b}) \times\widehat{i}) \cdot\widehat{k}=\frac{23}{2}, then ∣b→×2j^∣|\overrightarrow{b}\times 2\widehat{j}| is equal to

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  5. Question 5

    29 Jul 2022

    70
    Let a→=3i^+j^\overrightarrow{a}= 3\widehat{i}+\widehat{j} and b→=i^+2j^+k^\overrightarrow{b}=\widehat{i}+ 2\widehat{j}+\widehat{k}. Let c→\overrightarrow{c} be a vector satisfying a→×(b→×c→)=b→+λc→\overrightarrow{a}\times (\overrightarrow{b}\times\overrightarrow{c}) =\overrightarrow{b}+ \lambda\overrightarrow{c}. If b→\overrightarrow{b} and c→\overrightarrow{c} are non-parallel, then the value of λ\lambda is:

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  6. Question 6

    25 Jan 2023

    63
    Let a→,b→\overrightarrow{a},\overrightarrow{b} and c→\overrightarrow{c} be three non zero vectors such that b→⋅c→=0\overrightarrow{b}\cdot\overrightarrow{c}= 0 and a→×(b→×c→)=b→−c→2\overrightarrow{a}\times (\overrightarrow{b}\times\overrightarrow{c}) =\frac{\overrightarrow{b}-\overrightarrow{c}}{2}. If d→\overrightarrow{d} be a vector such that b→⋅d→=a→⋅b→\overrightarrow{b}\cdot\overrightarrow{d}=\overrightarrow{a}\cdot\overrightarrow{b}, then (a→×b→)⋅(c→×d→)(\overrightarrow{a}\times\overrightarrow{b}) \cdot (\overrightarrow{c}\times\overrightarrow{d}) is equal to

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  7. Question 7

    30 Jan 2023

    64
    If a→,b→,c→\overrightarrow{a},\overrightarrow{b},\overrightarrow{c} are three non-zero vectors and n^\widehat{n} is a unit vector perpendicular to c→\overrightarrow{c} such that a→=αb→−n^,(α≠\overrightarrow{a}= \alpha\overrightarrow{b}-\widehat{n},(\alpha \neq 0)0) and b→⋅c→=12\overrightarrow{b}\cdot\overrightarrow{c}= 12, then ∣c→×(a→×b→)∣|\overrightarrow{c}\times (\overrightarrow{a}\times\overrightarrow{b})| is equal to :

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  8. Question 8

    30 Jan 2023

    154
    Let λ∈R,a→=λı^+2ȷ^−3k^,b→=ı^−λȷ^+2k^\lambda \in R,\overrightarrow{a}= \lambda\widehat{\imath}+ 2\widehat{\jmath}- 3\widehat{k},\overrightarrow{b}=\widehat{\imath}- \lambda\widehat{\jmath}+ 2\widehat{k} If ((a→+b→)×(a→×b→))×(a→−b→)=8ı^−40ȷ^−24k^((\overrightarrow{a}+\overrightarrow{b}) \times (\overrightarrow{a}\times\overrightarrow{b})) \times (\overrightarrow{a}-\overrightarrow{b}) = 8\widehat{\imath}- 40\widehat{\jmath}- 24\widehat{k}, then ∣λ(a→+b→)×(a→−b→)∣2|\lambda(\overrightarrow{a}+\overrightarrow{b}) \times (\overrightarrow{a}-\overrightarrow{b})|^{2} is equal to

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  9. Question 9

    1 Feb 2024

    68
    Let a→=−5i^+j^−3k^,b→=i^+2j^−4k^\overrightarrow{a}= - 5\widehat{i}+\widehat{j}- 3\widehat{k},\overrightarrow{b}=\widehat{i}+ 2\widehat{j}- 4\widehat{k} and c→=(((a→×b→)×i^)×i^)×i^\overrightarrow{c}= (((\overrightarrow{a}\times\overrightarrow{b}) \times\widehat{i}) \times\widehat{i}) \times\widehat{i}. Then c→⋅(−i^+j^+k^)\overrightarrow{c}\cdot ( -\widehat{i}+\widehat{j}+\widehat{k}) is equal to

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  10. Question 10

    6 Apr 2024

    167
    Let a→=2i^+j^−k^,b→=((a→×(i^+j^))×i^)×i^\overrightarrow{a}= 2\widehat{i}+\widehat{j}-\widehat{k},\overrightarrow{b}= ((\overrightarrow{a}\times (\widehat{i}+\widehat{j})) \times\widehat{i}) \times\widehat{i}. Then the square of the projection of a→\overrightarrow{a} on b→\overrightarrow{b} is:

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  11. Question 11

    23 Jan 2026 S1

    53
    Let a→=−i^+j^+2k^,b→=i^−j^−3k^,c→=a→×b→\overrightarrow{a}= -\widehat{i}+\widehat{j}+ 2\widehat{k},\overrightarrow{b}=\widehat{i}-\widehat{j}- 3\widehat{k},\overrightarrow{c}=\overrightarrow{a}\times\overrightarrow{b} and d→=c→×a→\overrightarrow{d}=\overrightarrow{c}\times\overrightarrow{a}. Then (a→−b→)⋅d→(\overrightarrow{a}-\overrightarrow{b}) \cdot\overrightarrow{d} is equal to :

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