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Triangle with side vectors: evaluate dot and cross product expressions

This type of question, asked 2 times: 25 Jul 2022, 21 Jan 2026 S2. All 2 questions below, oldest first.

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  1. Question 1

    25 Jul 2022

    75
    Let ABCABC be a triangle such that BC→=a→,CA→=b→\overrightarrow{BC}=\overrightarrow{a},\overrightarrow{CA}=\overrightarrow{b}, AB→=c→,∣a→∣=62,∣b→∣=23\overrightarrow{AB}=\overrightarrow{c},|\overrightarrow{a}| = 6\sqrt{2},|\overrightarrow{b}| = 2\sqrt{3} and b→⋅c→=12\overrightarrow{b}\cdot\overrightarrow{c}= 12 Consider the statements : (S1): ∣(a→×b→)+(c→×b→)∣−∣c→∣=6(22−1)|(\overrightarrow{a}\times\overrightarrow{b}) + (\overrightarrow{c}\times\overrightarrow{b})| - |\overrightarrow{c}| = 6(2\sqrt{2}- 1) (S2): ∠ABC=cos⁡−1(23)\angle ABC =\cos^{- 1}\left( \sqrt{\frac{2}{3}} \right). Then

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  2. Question 2

    21 Jan 2026 S2

    65
    For a triangle ABCABC, let p→=BC→,q→=CA→\overrightarrow{p}=\overrightarrow{BC},\overrightarrow{q}=\overrightarrow{CA} and r→=BA→\overrightarrow{r}=\overrightarrow{BA}. If ∣p→∣=23,∣q→∣=2|\overrightarrow{p}| = 2\sqrt{3},|\overrightarrow{q}| = 2 and cos⁡θ=13\cos\theta =\frac{1}{\sqrt{3}}, where θ\theta is the angle between p→\overrightarrow{p} and q→\overrightarrow{q}, then ∣p→×(q→−3r→)∣2+3∣r→∣2|\overrightarrow{p}\times (\overrightarrow{q}- 3\overrightarrow{r})|^{2}+ 3|\overrightarrow{r}|^{2} is equal to:

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