PYQ Vault
EbooksTutor

Find an unknown in a vector from the size of a cross product (area of a parallelogram or triangle)

This type of question, asked 7 times: 2021 Paper 4, 2021 Paper 20, 28 Jun 2022, 13 Apr 2023, 4 Apr 2024, 9 Apr 2024, 22 Jan 2026 S2. All 7 questions below, oldest first.

Try each one first, then open its answer.

  1. Question 1

    2021 Paper 4

    89
    Let a→=i^+αj^+3k^\overrightarrow{a}=\widehat{i}+\alpha\widehat{j}+ 3\widehat{k} and b→=3i^−αj^+k^\overrightarrow{b}= 3\widehat{i}-\alpha\widehat{j}+\widehat{k}. If the area of the parallelogram whose adjacent sides are represented by the vectors a→\overrightarrow{a} and b→\overrightarrow{b} is 838\sqrt{3} square units, then a→⋅b→\overrightarrow{a}\cdot\overrightarrow{b} is equal to

  2. Question 2

    2021 Paper 20

    81
    Let a→=i^+5j^+αk^,b→=i^+3j^+βk^\overrightarrow{a}=\widehat{i}+ 5\widehat{j}+\alpha\widehat{k},\overrightarrow{b}=\widehat{i}+ 3\widehat{j}+\beta\widehat{k} and c→=−i^+2j^−3k^\overrightarrow{c}= -\widehat{i}+ 2\widehat{j}- 3\widehat{k} be three vectors such that, ∣b→×c→∣=53|\overrightarrow{b}\times\overrightarrow{c}| = 5\sqrt{3} and a→\overrightarrow{a} is perpendicular to b→\overrightarrow{b}. Then the greatest amongst the values of ∣a→∣2|\overrightarrow{a}|^{2} is

  3. Question 3

    28 Jun 2022

    164
    Let a→=αi^+2j^−k^\overrightarrow{a}= \alpha\widehat{i}+ 2\widehat{j}-\widehat{k} and b→=−2i^+αj^+k^\overrightarrow{b}= - 2\widehat{i}+ \alpha\widehat{j}+\widehat{k}, where α∈R\alpha \in R. If the area of the parallelogram whose adjacent sides are represented by the vectors a→\overrightarrow{a} and b→\overrightarrow{b} is 15(α2+4)\sqrt{15\left( \alpha^{2}+ 4 \right)}, then the value of 2∣a→∣2+(a→⋅b→)∣b→∣22|\overrightarrow{a}|^{2}+ (\overrightarrow{a}\cdot\overrightarrow{b})|\overrightarrow{b}|^{2} is equal to

    Tap an option to check your answer.

  4. Question 4

    13 Apr 2023

    168
    Let for a triangle ABCABC,
    AB→=−2i^+j^+3k^\overrightarrow{AB}= - 2\widehat{i}+\widehat{j}+ 3\widehat{k}
    CB→=αi^+βj^+γk^\overrightarrow{CB}= \alpha\widehat{i}+ \beta\widehat{j}+ \gamma\widehat{k}
    CA→=4i^+3j^+δk^\overrightarrow{CA}= 4\widehat{i}+ 3\widehat{j}+ \delta\widehat{k}
    If δ>0\delta > 0 and the area of the triangle ABCABC is 565\sqrt{6}, then CB→⋅CA→\overrightarrow{CB}\cdot\overrightarrow{CA} is equal to

    Tap an option to check your answer.

  5. Question 5

    4 Apr 2024

    89
    Let ABCABC be a triangle of area 15215\sqrt{2} and the vectors AB→=i^+2j^−7k^, BC→=ai^+bj^+ck^\overrightarrow{AB}=\widehat{i}+ 2\widehat{j}- 7\widehat{k},\ \overrightarrow{BC}= a\widehat{i}+b\widehat{j}+c\widehat{k} and AC→=6i^+dj^−2k^,d>0\overrightarrow{AC}= 6\widehat{i}+ d\widehat{j}- 2\widehat{k},d> 0. Then the square of the length of the largest side of the triangle ABCABC is

  6. Question 6

    9 Apr 2024

    78
    Let three vectors a→=αi^+4j^+2k^\overrightarrow{a}= \alpha\widehat{i}+ 4\widehat{j}+ 2\widehat{k}, b→=5i^+3j^+4k^,c→=xi^+yj^+zk^\overrightarrow{b}= 5\widehat{i}+ 3\widehat{j}+ 4\widehat{k},\overrightarrow{c}= x\widehat{i}+ y\widehat{j}+ z\widehat{k} from a triangle such that c→=a→−b→\overrightarrow{c}=\overrightarrow{a}-\overrightarrow{b} and the area of the triangle is 565\sqrt{6}. if α\alpha is a positive real number, then ∣c→∣2|\overrightarrow{c}|^{2} is:

    Tap an option to check your answer.

  7. Question 7

    22 Jan 2026 S2

    58
    Let a→=2i^−j^+k^\overrightarrow{a}= 2\widehat{i}-\widehat{j}+\widehat{k} and b→=λj^+2k^,λ∈Z\overrightarrow{b}= \lambda\widehat{j}+ 2\widehat{k},\lambda \in Z be two vectors, Let c→=a→×b→\overrightarrow{c}=\overrightarrow{a}\times\overrightarrow{b} and d→\overrightarrow{d} be a vector of magnitude 2 in yz-plane. If ∣c→∣=53|\overrightarrow{c}| =\sqrt{53}, then the maximum possible value of (c→⋅d→)2(\overrightarrow{c}\cdot\overrightarrow{d})^{2} is equal to :

    Tap an option to check your answer.