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Cross product of combinations: (pa + qb) x (ra + sb) = (ps - qr)(a x b)

This type of question, asked 4 times: 13 Apr 2023, 29 Jan 2024, 9 Apr 2024, 6 Apr 2026 S2. All 4 questions below, oldest first.

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  1. Question 1

    13 Apr 2023

    157
    Let ∣a→∣=2,∣b→∣=3|\overrightarrow{a}| = 2,|\overrightarrow{b}| = 3 and the angle between the vectors a→\overrightarrow{a} and b→\overrightarrow{b} be π4\frac{\pi}{4}. Then ∣(a→+2b→)×(2a→−3b→)∣2|(\overrightarrow{a}+ 2\overrightarrow{b}) \times (2\overrightarrow{a}- 3\overrightarrow{b})|^{2} is equal to

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  2. Question 2

    29 Jan 2024

    158
    Let OA→=a→,OB→=12a→+4b→\overrightarrow{OA}=\overrightarrow{a},\overrightarrow{OB}= 12\overrightarrow{a}+ 4\overrightarrow{b} and OC→=b→\overrightarrow{OC}=\overrightarrow{b}, where OO is the origin. If SS is the parallelogram with adjacent sides OAOA and OCOC, then  area of the quadrilateral OABC area of S\frac{\text{~area of the quadrilateral~}OABC}{\text{~area of~}S} is equal to

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  3. Question 3

    9 Apr 2024

    70
    Let OA→=2a→,OB→=6a→+5b→\overrightarrow{OA}= 2\overrightarrow{a},\overrightarrow{OB}= 6\overrightarrow{a}+ 5\overrightarrow{b} and OC→=3b→\overrightarrow{OC}= 3\overrightarrow{b}, where OO is the origin. If the area of the parallelogram with adjacent sides OA→\overrightarrow{OA} and OC→\overrightarrow{OC} is 15 sq. units, then the area (in sq. units) of the quadrilateral OABCOABC is equal to:

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  4. Question 4

    6 Apr 2026 S2

    65
    Let a→=2i^+3j^+3k^\overrightarrow{a}= 2\widehat{i}+ 3\widehat{j}+ 3\widehat{k} and b→=6i^+3j^+3k^\overrightarrow{b}= 6\widehat{i}+ 3\widehat{j}+ 3\widehat{k}. Then the square of the area of the triangle with adjacent sides determined by the vectors (2a→+3b→)(2\overrightarrow{a}+ 3\overrightarrow{b}) and (a→−b→)(\overrightarrow{a}-\overrightarrow{b}) is :

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