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Vector equations like r x a = c x a or p(a x c) + q(b x c) = 0: the vector is parallel to a known one, a dot condition fixes it

This type of question, asked 18 times: 2021 Paper 3, 2021 Paper 6, 2021 Paper 7, 24 Jan 2023, 29 Jan 2023, 31 Jan 2023, 1 Feb 2023, 8 Apr 2023, 13 Apr 2023, 31 Jan 2024, 1 Feb 2024, 5 Apr 2024, 8 Apr 2024, 29 Jan 2025, 3 Apr 2025, 24 Jan 2026 S2, 5 Apr 2026 S1, 8 Apr 2026 S2. All 21 questions below, oldest first.

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  1. Question 1

    2021 Paper 3

    88
    Let a→=i^+2j^−k^,b→=i^−j^\overrightarrow{a}=\widehat{i}+ 2\widehat{j}-\widehat{k},\overrightarrow{b}=\widehat{i}-\widehat{j} and c→=i^−j^−k^\overrightarrow{c}=\widehat{i}-\widehat{j}-\widehat{k} be three given vectors. If r→\overrightarrow{r} is a vector such that r→×a→=c→×a→\overrightarrow{r}\times\overrightarrow{a}=\overrightarrow{c}\times\overrightarrow{a} and r→⋅b→=0\overrightarrow{r}\cdot\overrightarrow{b}= 0, then r→⋅a→\overrightarrow{r}\cdot\overrightarrow{a} is equal to

  2. Question 2

    2021 Paper 6

    80
    Let a→=i^+2j^−3k^\overrightarrow{a}=\widehat{i}+ 2\widehat{j}- 3\widehat{k} and b→=2i^−3j^+5k^\overrightarrow{b}= 2\widehat{i}- 3\widehat{j}+ 5\widehat{k}. If r→×a→=b→×r→,r→⋅(αi^+2j^+k^)=3\overrightarrow{r}\times\overrightarrow{a}=\overrightarrow{b}\times\overrightarrow{r},\overrightarrow{r}\cdot (\alpha\widehat{i}+ 2\widehat{j}+\widehat{k}) = 3 and r→⋅(2i^+5j^−αk^)=−1,α∈R\overrightarrow{r}\cdot (2\widehat{i}+ 5\widehat{j}-\alpha\widehat{k}) = - 1,\alpha\in R, then the value of α+∣r→∣2\alpha+ |\overrightarrow{r}|^{2} is equal to:

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  3. Question 3

    2021 Paper 7

    62
    Let a→=2i^−3j^+4k^\overrightarrow{a}= 2\widehat{i}- 3\widehat{j}+ 4\widehat{k} and b→=7i^+j^−6k^\overrightarrow{b}= 7\widehat{i}+\widehat{j}- 6\widehat{k}. If r→×a→=r→×b→,r→⋅(i^+2j^+k^)=−3\overrightarrow{r}\times\overrightarrow{a}=\overrightarrow{r}\times\overrightarrow{b},\overrightarrow{r}\cdot (\widehat{i}+ 2\widehat{j}+\widehat{k}) = - 3, then r→⋅(2i^−3j^+k^)\overrightarrow{r}\cdot (2\widehat{i}- 3\widehat{j}+\widehat{k}) is equal to :

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  4. Question 4

    24 Jan 2023

    177
    Let a→=i^+2j^+λk^,b→=3i^−5j^−λk^,a→⋅c→=7,2b→⋅c→+43=0,a→×c→=b→×c→\overrightarrow{a}=\widehat{i}+ 2\widehat{j}+ \lambda\widehat{k},\overrightarrow{b}= 3\widehat{i}- 5\widehat{j}- \lambda\widehat{k},\overrightarrow{a}\cdot\overrightarrow{c}= 7,2\overrightarrow{b}\cdot\overrightarrow{c}+ 43 = 0,\overrightarrow{a}\times\overrightarrow{c}=\overrightarrow{b}\times\overrightarrow{c}. Then ∣a→⋅b→∣|\overrightarrow{a}\cdot\overrightarrow{b}| is equal to

  5. Question 5

    29 Jan 2023

    156
    If a→=ı^+2k^,b→=ı^+ȷ^+k^,c→=7ı^−3ȷ^+4k^,r→×b→+b→×c→=0→\overrightarrow{a}=\widehat{\imath}+ 2\widehat{k},\overrightarrow{b}=\widehat{\imath}+\widehat{\jmath}+\widehat{k},\overrightarrow{c}= 7\widehat{\imath}- 3\widehat{\jmath}+ 4\widehat{k},\overrightarrow{r}\times\overrightarrow{b}+\overrightarrow{b}\times\overrightarrow{c}=\overrightarrow{0} and r→⋅a→=0\overrightarrow{r}\cdot\overrightarrow{a}= 0. Then r→⋅c→\overrightarrow{r}\cdot\overrightarrow{c} is equal to

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  6. Question 6

    31 Jan 2023

    164
    Let: a→=ı^+2ȷ^+3k^,b→=ı^−ȷ^+2k^\overrightarrow{a}=\widehat{\imath}+ 2\widehat{\jmath}+ 3\widehat{k},\overrightarrow{b}=\widehat{\imath}-\widehat{\jmath}+ 2\widehat{k} and c→=5ı^−3ȷ^+3k^\overrightarrow{c}= 5\widehat{\imath}- 3\widehat{\jmath}+ 3\widehat{k} be there vectors. If r→\overrightarrow{r} is a vector such that, r→×b→=c→×b→\overrightarrow{r}\times\overrightarrow{b}=\overrightarrow{c}\times\overrightarrow{b} and r→⋅a→=0\overrightarrow{r}\cdot\overrightarrow{a}= 0, then 25∣r→∣225|\overrightarrow{r}|^{2} is equal to

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  7. Question 7

    1 Feb 2023

    154
    Let a→=2i^−7j^+5k^,b→=i^+k^\overrightarrow{a}= 2\widehat{i}- 7\widehat{j}+ 5\widehat{k},\overrightarrow{b}=\widehat{i}+\widehat{k} and c→=i^+2j^−3k^\overrightarrow{c}=\widehat{i}+ 2\widehat{j}- 3\widehat{k} be three given vectors. If r→\overrightarrow{r} is a vector such that r→×a→=c→×a→\overrightarrow{r}\times\overrightarrow{a}=\overrightarrow{c}\times\overrightarrow{a} and r→⋅b→=0\overrightarrow{r}\cdot\overrightarrow{b}= 0, then ∣r→∣|\overrightarrow{r}| is equal to:

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  8. Question 8

    8 Apr 2023

    83
    Let a→=6i^+9j^+12k^,b→=αi^+11j^−2k^\overrightarrow{a}= 6\widehat{i}+ 9\widehat{j}+ 12\widehat{k},\overrightarrow{b}= \alpha\widehat{i}+ 11\widehat{j}- 2\widehat{k} and c→\overrightarrow{c} be vectors such that a→×c→=a→×b→\overrightarrow{a}\times\overrightarrow{c}=\overrightarrow{a}\times\overrightarrow{b}. If a→⋅c→=−12\overrightarrow{a}\cdot\overrightarrow{c}= - 12, c→⋅(i^−2j^+k^)=5\overrightarrow{c}\cdot (\widehat{i}- 2\widehat{j}+\widehat{k}) = 5, then c→⋅(i^+j^+k^)\overrightarrow{c}\cdot (\widehat{i}+\widehat{j}+\widehat{k}) is equal to

  9. Question 9

    13 Apr 2023

    76
    Let a=i^+4j^+2k^,b=3i^−2j^+7k^a =\widehat{i}+ 4\widehat{j}+ 2\widehat{k},b = 3\widehat{i}- 2\widehat{j}+ 7\widehat{k} and c→=2i^−j^+4k^\overrightarrow{c}= 2\widehat{i}-\widehat{j}+ 4\widehat{k}. If a vector d→\overrightarrow{d} satisfies d→×b→=c→×b→\overrightarrow{d}\times\overrightarrow{b}=\overrightarrow{c}\times\overrightarrow{b} and d→.a→=24\overrightarrow{d}.\overrightarrow{a}= 24, then ∣d→∣2|\overrightarrow{d}|^{2} is equal to

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  10. Question 10

    31 Jan 2024

    72
    Let  a→=3i^+j^−2k^, b→=4i^+j^+7k^\ \overrightarrow{a}= 3\widehat{i}+\widehat{j}- 2\widehat{k},\ \overrightarrow{b}= 4\widehat{i}+\widehat{j}+ 7\widehat{k} and c→=i^−3j^+4k^\overrightarrow{c}=\widehat{i}- 3\widehat{j}+ 4\widehat{k} be three vectors. If a vectors p→\overrightarrow{p} satisfies p→×b→=c→×b→\overrightarrow{p}\times\overrightarrow{b}=\overrightarrow{c}\times\overrightarrow{b} and p→⋅a→=0\overrightarrow{p}\cdot\overrightarrow{a}= 0, then p→⋅(i^−j^−k^)\overrightarrow{p}\cdot (\widehat{i}-\widehat{j}-\widehat{k}) is equal to

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  11. Question 11

    1 Feb 2024

    175
    Let  a→=i^+j^+k^, b→=−i^−8j^+2k^\ \overrightarrow{a}=\widehat{i}+\widehat{j}+\widehat{k},\ \overrightarrow{b}= -\widehat{i}- 8\widehat{j}+ 2\widehat{k} and c→=4i^+c2j^+c3k^\overrightarrow{c}= 4\widehat{i}+c_{2}\widehat{j}+c_{3}\widehat{k} be three vectors such that b→×a→=c→×a→\overrightarrow{b}\times\overrightarrow{a}=\overrightarrow{c}\times\overrightarrow{a}. If the angle between the vector c→\overrightarrow{c} and the vector 3i^+4j^+k^3\widehat{i}+ 4\widehat{j}+\widehat{k} is θ\theta, then the greatest integer less than or equal to tan⁡2θ\tan^{2}\theta is:

  12. Question 12

    5 Apr 2024

    88
    Let a→=i^−3j^+7k^, b→=2i^−j^+k^\overrightarrow{a}=\widehat{i}- 3\widehat{j}+ 7\widehat{k},\ \overrightarrow{b}= 2\widehat{i}-\widehat{j}+\widehat{k} and c→\overrightarrow{c} be a vector such that (a→+2b→)×c→=3(c→×a→)(\overrightarrow{a}+ 2\overrightarrow{b}) \times\overrightarrow{c}= 3(\overrightarrow{c}\times\overrightarrow{a}). If a→⋅c→=130\overrightarrow{a}\cdot\overrightarrow{c}= 130, then b→⋅c→\overrightarrow{b}\cdot\overrightarrow{c} is equal to

  13. Question 13

    5 Apr 2024

    156
    Let a→=2i^+5j^−k^,b→=2i^−2j^+2k^\overrightarrow{a}= 2\widehat{i}+ 5\widehat{j}-\widehat{k},\overrightarrow{b}= 2\widehat{i}- 2\widehat{j}+ 2\widehat{k} and c→\overrightarrow{c} be three vectors such that
    (c→+i^)×(a→+b→+i^)=a→×(c→+i^)(\overrightarrow{c}+\widehat{i}) \times (\overrightarrow{a}+\overrightarrow{b}+\widehat{i}) =\overrightarrow{a}\times (\overrightarrow{c}+\widehat{i})
    and a→⋅c→=−29\overrightarrow{a}\cdot\overrightarrow{c}= - 29, then c→⋅(−2i^+j^+k^)\overrightarrow{c}\cdot ( - 2\widehat{i}+\widehat{j}+\widehat{k}) is equal to:

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  14. Question 14

    8 Apr 2024

    88
    Let a→=9i^−13j^+25k^,b→=3i^+7j^−13k^\overrightarrow{a}= 9\widehat{i}- 13\widehat{j}+ 25\widehat{k},\overrightarrow{b}= 3\widehat{i}+ 7\widehat{j}- 13\widehat{k} and c→=17i^−2j^+k^\overrightarrow{c}= 17\widehat{i}- 2\widehat{j}+\widehat{k} be three given vectors. If r→\overrightarrow{r} is a vector such that r→×a→=(b→+c→)×a→\overrightarrow{r}\times\overrightarrow{a}= (\overrightarrow{b}+\overrightarrow{c}) \times\overrightarrow{a} and r→⋅(b→−c→)=0\overrightarrow{r}\cdot (\overrightarrow{b}-\overrightarrow{c}) = 0, then ∣593r→+67a→∣2(593)2\frac{|593\overrightarrow{r}+ 67\overrightarrow{a}|^{2}}{(593)^{2}} is equal to

  15. Question 15

    8 Apr 2024

    163
    Let a→=4i^−j^+k^,b→=11i^−j^+k^\overrightarrow{a}= 4\widehat{i}-\widehat{j}+\widehat{k},\overrightarrow{b}= 11\widehat{i}-\widehat{j}+\widehat{k} and c→\overrightarrow{c} be a vector such that (a→+b→)×c→=c→×(−2a→+3b→)(\overrightarrow{a}+\overrightarrow{b}) \times\overrightarrow{c}=\overrightarrow{c}\times ( - 2\overrightarrow{a}+ 3\overrightarrow{b}). If (2a→+3b→)⋅c→=1670(2\overrightarrow{a}+ 3\overrightarrow{b}) \cdot\overrightarrow{c}= 1670, then ∣c→∣2|\overrightarrow{c}|^{2} is equal to:

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  16. Question 16

    29 Jan 2025

    55
    Let a→=2i^−j^+3k^,b→=3i^−5j^+k^\overrightarrow{a}= 2\widehat{i}-\widehat{j}+ 3\widehat{k},\overrightarrow{b}= 3\widehat{i}- 5\widehat{j}+\widehat{k} and c→\overrightarrow{c} be a vector such that a→×c→=c→×b→\overrightarrow{a}\times\overrightarrow{c}=\overrightarrow{c}\times\overrightarrow{b} and (a→+c→)⋅(b→+c→)=168(\overrightarrow{a}+\overrightarrow{c}) \cdot (\overrightarrow{b}+\overrightarrow{c}) = 168. Then the maximum value of ∣c→∣2|\overrightarrow{c}|^{2} is :

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  17. Question 17

    3 Apr 2025

    73
    Let a→=i^+j^+k^,b→=3i^+2j^−k^,c→=λj^+μk^\overrightarrow{a} = \widehat{i} + \widehat{j} + \widehat{k},\overrightarrow{b} = 3\widehat{i} + 2\widehat{j} - \widehat{k},\overrightarrow{c} = \lambda\widehat{j} + \mu\widehat{k} and d^\widehat{d} be a unit vector such that a→×d^=b→×d^\overrightarrow{a} \times \widehat{d} = \overrightarrow{b} \times \widehat{d} and c→.d^=1\overrightarrow{c}.\widehat{d} = 1, If c→\overrightarrow{c} is perpendicular to a→\overrightarrow{a}, then ∣3λd^+μc→∣2|3\lambda\widehat{d} + \mu\overrightarrow{c}|^{2} is equal to

  18. Question 18

    3 Apr 2025

    149
    Let a→=i^+2j^+k^,b→=3i^−3j^+3k^\overrightarrow{a} = \widehat{i} + 2\widehat{j} + \widehat{k},\overrightarrow{b} = 3\widehat{i} - 3\widehat{j} + 3\widehat{k}, c→=2i^−j^+2k^\overrightarrow{c} = 2\widehat{i} - \widehat{j} + 2\widehat{k} and d→\overrightarrow{d} be a vector such that b→×d→=c→×d→\overrightarrow{b} \times \overrightarrow{d} = \overrightarrow{c} \times \overrightarrow{d} and a→⋅d→=4\overrightarrow{a} \cdot \overrightarrow{d} = 4. Then ∣(a→×d→)∣2|(\overrightarrow{a} \times \overrightarrow{d})|^{2} is equal to.

  19. Question 19

    24 Jan 2026 S2

    69
    Let a→=2i^−5j^+5k^\overrightarrow{a}= 2\widehat{i}- 5\widehat{j}+ 5\widehat{k} and b→=i^−j^+3k^\overrightarrow{b}=\widehat{i}-\widehat{j}+ 3\widehat{k}. If c→\overrightarrow{c} is a vector such that 2(a→×c→)+3(b→×c→)=0→2(\overrightarrow{a}\times\overrightarrow{c}) + 3(\overrightarrow{b}\times\overrightarrow{c}) =\overrightarrow{0} and (a→−b→)⋅c→=−97(\overrightarrow{a}-\overrightarrow{b}) \cdot\overrightarrow{c}= - 97, then ∣c→×k^∣2|\overrightarrow{c}\times\widehat{k}|^{2} is equal to

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  20. Question 20

    5 Apr 2026 S1

    64
    Let a→=7i^+j^−k^\overrightarrow{a}=\sqrt{7}\widehat{i}+\widehat{j}-\widehat{k} and b→=j^+2k^\overrightarrow{b}=\widehat{j}+ 2\widehat{k}. If r→\overrightarrow{r} is a vector such that r→×a→+a→×b→=0→\overrightarrow{r}\times\overrightarrow{a}+\overrightarrow{a}\times\overrightarrow{b}=\overrightarrow{0} and r→⋅a→=0\overrightarrow{r}\cdot\overrightarrow{a}= 0, then ∣3r→∣2|3\overrightarrow{r}|^{2} is equal to :

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  21. Question 21

    8 Apr 2026 S2

    64
    Let a→=4i^−j^+3k^,b→=10i^+2j^−k^\overrightarrow{a}= 4\widehat{i}-\widehat{j}+ 3\widehat{k},\overrightarrow{b}= 10\widehat{i}+ 2\widehat{j}-\widehat{k} and a vector c→\overrightarrow{c} be such that 2(a→×b→)+3(b→×c→)=0→2(\overrightarrow{a}\times\overrightarrow{b}) + 3(\overrightarrow{b}\times\overrightarrow{c}) =\overrightarrow{0}. If a→⋅c→=15\overrightarrow{a}\cdot\overrightarrow{c}= 15, then c→⋅(i^+j^−3k^)\overrightarrow{c}\cdot (\widehat{i}+\widehat{j}- 3\widehat{k}) is equal to :

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