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Find a vector from its cross product and dot product with a known vector

This type of question, asked 13 times: 2021 Paper 18, 25 Jun 2022, 28 Jun 2022, 29 Jun 2022, 25 Jul 2022, 24 Jan 2023, 25 Jan 2023, 27 Jan 2024, 31 Jan 2024, 6 Apr 2024, 2 Apr 2025, 21 Jan 2026 S1, 6 Apr 2026 S1. All 13 questions below, oldest first.

Try each one first, then open its answer.

  1. Question 1

    2021 Paper 18

    72
    Let a→=i^+j^+k^\overrightarrow{a}=\widehat{i}+\widehat{j}+\widehat{k} and b→=j^−k^\overrightarrow{b}=\widehat{j}-\widehat{k}. If c→\overrightarrow{c} is a vector such that a→×c→=b→\overrightarrow{a}\times\overrightarrow{c}=\overrightarrow{b} and a→.c→=3\overrightarrow{a}.\overrightarrow{c}= 3, then a→⋅(b→×c→)\overrightarrow{a}\cdot (\overrightarrow{b}\times\overrightarrow{c}) is equal to:

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  2. Question 2

    25 Jun 2022

    176
    Let b→=i^+j^+λk^,λ∈R\overrightarrow{b}=\widehat{i}+\widehat{j}+ \lambda\widehat{k},\lambda \in R. If a→\overrightarrow{a} is a vector such that a→×b→=13i^−j^−4k^\overrightarrow{a}\times\overrightarrow{b}= 13\widehat{i}-\widehat{j}- 4\widehat{k} and  a→⋅b→+21=0,\ \overrightarrow{a}\cdot\overrightarrow{b}+ 21 = 0, then (b→−a→)⋅(k^−j^)+(b→+a→)⋅(i^−k^)(\overrightarrow{b}-\overrightarrow{a}) \cdot (\widehat{k}-\widehat{j}) + (\overrightarrow{b}+\overrightarrow{a}) \cdot (\widehat{i}-\widehat{k}) is equal to

  3. Question 3

    28 Jun 2022

    169
    Let a→\overrightarrow{a} be a vector which is perpendicular to the vector 3i^+12j^+2k^3\widehat{i}+\frac{1}{2}\widehat{j}+ 2\widehat{k}. If a→×(2i^+k^)=2i^−13j^−4k^\overrightarrow{a}\times (2\widehat{i}+\widehat{k}) = 2\widehat{i}- 13\widehat{j}- 4\widehat{k}, then the projection of the vector a→\overrightarrow{a} on the vector 2i^+2j^+k^2\widehat{i}+ 2\widehat{j}+\widehat{k} is

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  4. Question 4

    29 Jun 2022

    171
    Let a→=i^−2j^+3k^,b→=i^+j^+k^\overrightarrow{a}=\widehat{i}- 2\widehat{j}+ 3\widehat{k},\overrightarrow{b}=\widehat{i}+\widehat{j}+\widehat{k} and c→\overrightarrow{c} be a vector such that a→+(b→×c→)=0→\overrightarrow{a}+ (\overrightarrow{b}\times\overrightarrow{c}) =\overrightarrow{0} and b→⋅c→=5\overrightarrow{b}\cdot\overrightarrow{c}= 5. Then, the value of 3(c→⋅a→)3(\overrightarrow{c}\cdot\overrightarrow{a}) is equal to

  5. Question 5

    25 Jul 2022

    167
    Let a→=i^−j^+2k^\overrightarrow{a}=\widehat{i}-\widehat{j}+ 2\widehat{k} and b→\overrightarrow{b} be a vector such that a→×b→=2i^−k^\overrightarrow{a}\times\overrightarrow{b}= 2\widehat{i}-\widehat{k} and a→⋅b→=3\overrightarrow{a}\cdot\overrightarrow{b}= 3. Then the projection of b→\overrightarrow{b} on the vector a→−b→\overrightarrow{a}-\overrightarrow{b} is:-

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  6. Question 6

    24 Jan 2023

    61
    Let u→=ı^−ȷ^−2k^,v→=2ı^+ȷ^−k^,v→⋅w→=2\overrightarrow{u}=\widehat{\imath}-\widehat{\jmath}- 2\widehat{k},\overrightarrow{v}= 2\widehat{\imath}+\widehat{\jmath}-\widehat{k},\overrightarrow{v}\cdot\overrightarrow{w}= 2 and v→×w→=u→+λv→\overrightarrow{v}\times\overrightarrow{w}=\overrightarrow{u}+ \lambda\overrightarrow{v}. Then u→⋅w→\overrightarrow{u}\cdot\overrightarrow{w} is equal to

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  7. Question 7

    25 Jan 2023

    159
    Let a→=−i^−j^+k^,a→⋅b→=1\overrightarrow{a}= -\widehat{i}-\widehat{j}+\widehat{k},\overrightarrow{a}\cdot\overrightarrow{b}= 1 and a→×b→=i^−j^\overrightarrow{a}\times\overrightarrow{b}=\widehat{i}-\widehat{j}. Then a→−6b→\overrightarrow{a}- 6\overrightarrow{b} is equal to

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  8. Question 8

    27 Jan 2024

    77
    Let a→=i^+2j^+k,b→=3(i^−j^+k)\overrightarrow{a}=\widehat{i}+ 2\widehat{j}+ k,\overrightarrow{b}= 3(\widehat{i}-\widehat{j}+ k). Let c→\overrightarrow{c} be the vector such that a→×c→=b→\overrightarrow{a}\times\overrightarrow{c}=\overrightarrow{b} and a→⋅c→=3\overrightarrow{a}\cdot\overrightarrow{c}= 3. Then a→⋅((c→×b→)−b→−c→)\overrightarrow{a}\cdot ((\overrightarrow{c}\times\overrightarrow{b}) -\overrightarrow{b}-\overrightarrow{c}) is equal to:

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  9. Question 9

    31 Jan 2024

    176
    Let a→=3i^+2j^+k^,b→=2i^−j^+3k^\overrightarrow{a}= 3\widehat{i}+ 2\widehat{j}+\widehat{k},\overrightarrow{b}= 2\widehat{i}-\widehat{j}+ 3\widehat{k} and c→\overrightarrow{c} be a vector such that (a→+b→)×c→=2(a→×b→)+24j^−6k^(\overrightarrow{a}+\overrightarrow{b}) \times\overrightarrow{c}= 2(\overrightarrow{a}\times\overrightarrow{b}) + 24\widehat{j}- 6\widehat{k} and (a→−b→+i^)⋅c→=−3(\overrightarrow{a}-\overrightarrow{b}+\widehat{i}) \cdot\overrightarrow{c}= - 3. Then ∣c→∣2|\overrightarrow{c}|^{2} is equal to

  10. Question 10

    6 Apr 2024

    90
    Let a→=2i^−3j^+4k^,b→=3i^+4j^−5k^\overrightarrow{a}= 2\widehat{i}- 3\widehat{j}+ 4\widehat{k},\overrightarrow{b}= 3\widehat{i}+ 4\widehat{j}- 5\widehat{k}, and a vector c→\overrightarrow{c} be such that a→×(b→+c→)+b→×c→=i^+8j^+13k^\overrightarrow{a}\times\left( \overrightarrow{b}+\overrightarrow{c} \right)+\overrightarrow{b}\times\overrightarrow{c}=\widehat{i}+ 8\widehat{j}+ 13\widehat{k}If a→⋅c→=13\overrightarrow{a}\cdot\overrightarrow{c}= 13, then (24−b→⋅c→)(24 -\overrightarrow{b}\cdot\overrightarrow{c}) is equal to

  11. Question 11

    2 Apr 2025

    143
    Let a→=2i^−3j^+k^,b→=3i^+2j^+5k^\overrightarrow{a}= 2\widehat{i}- 3\widehat{j}+\widehat{k},\overrightarrow{b}= 3\widehat{i}+ 2\widehat{j}+ 5\widehat{k} and a vector c→\overrightarrow{c} be such that (a→−c→)×b→=−18i^−3j^+12k^(\overrightarrow{a}-\overrightarrow{c}) \times\overrightarrow{b}= - 18\widehat{i}- 3\widehat{j}+ 12\widehat{k} and a→⋅c→=3\overrightarrow{a}\cdot\overrightarrow{c}= 3. If b→×c→=d→\overrightarrow{b}\times\overrightarrow{c}=\overrightarrow{d}, then ∣a→⋅d→∣|\overrightarrow{a}\cdot\overrightarrow{d}| is equal to :

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  12. Question 12

    21 Jan 2026 S1

    55
    Let a→=−i^+2j^+2k^,b→=8i^+7j^−3k^\overrightarrow{a}= -\widehat{i}+ 2\widehat{j}+ 2\widehat{k},\overrightarrow{b}= 8\widehat{i}+ 7\widehat{j}- 3\widehat{k} and c→\overrightarrow{c} be a vector such that a→×c→=b→\overrightarrow{a}\times\overrightarrow{c}=\overrightarrow{b}. If c→⋅(i^+j^+k^)=4\overrightarrow{c}\cdot (\widehat{i}+\widehat{j}+\widehat{k}) = 4, then ∣a→+c→∣2|\overrightarrow{a}+\overrightarrow{c}|^{2} is equal to :

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  13. Question 13

    6 Apr 2026 S1

    73
    If a→=i^+j^+k^,b→=j^−k^\overrightarrow{a} = \widehat{i} + \widehat{j} + \widehat{k},\overrightarrow{b} = \widehat{j} - \widehat{k} and c→\overrightarrow{c} be three vectors such that a→×c→=b→\overrightarrow{a} \times \overrightarrow{c} = \overrightarrow{b} and a→⋅c→=3\overrightarrow{a} \cdot \overrightarrow{c} = 3, then c→⋅(a→−2b→)\overrightarrow{c} \cdot (\overrightarrow{a} - 2\overrightarrow{b}) is equal to ____\_\_\_\_ .