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Let $\mathrm{ABC}$ be triangle such that $\overrightarrow{\mathrm{BC}}=\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{CA}}=\overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{c}},|\overrightarrow{\mathrm{a}}|=6 \sqrt{2},|\overrightarrow{\mathrm{b}}|=2 \sqrt{3}$ and $\vec{b} \cdot \vec{c}=12$ . Consider the statements :

$(\mathrm{S}1) :|(\vec{a} \times \vec{b})+(\vec{c} \times \vec{b})|-|\vec{c}|=6(2 \sqrt{2}-1)$

$(\mathrm{S} 2): \angle \mathrm{ACB}=\cos ^{-1}\left(\sqrt{\frac{2}{3}}\right)$

Then
Option: 1
both (S1) and (S2) are true

Option: 2
only (S1) is true

Option: 3
only (S2) is true

Option: 4
both (S1) and (S2) are false

Answers (1)

best_answer

$\vec{a}+\vec{b}+\vec{c}=0 \\$

$\vec{b}+\vec{c}=-\vec{a} \\$

$|\vec{b}|^{2}+|\vec{c}|^{2}+2 \vec{b} \cdot \vec{c}=|\vec{a}|^{2} \\$

$|\vec{c}|^{2}=36 \Rightarrow|\vec{c}|=6$

$\text { S1: }|\vec{a} \times \vec{b}+\vec{c} \times \vec{b}|-|\vec{c}| \\$

$\mid(\vec{a}+\vec{c}|\times \vec{b}|-|\vec{c}| \\$

$|-\vec{b} \times \vec{b}|-|\vec{c}| \\$

$0-6=-6 \\$

$S2: \vec{a}+\vec{b}+\vec{c}=0 \\$

$\vec{b}+\vec{c}=-\vec{a} \\$

$|\vec{a}|^{2}+|\vec{b}|^{2}-2|\vec{a}||\vec{b}| \cos (\angle A C B)=|\vec{c}|^{2} \\$

$\cos \mid \angle A C B)=\sqrt{\frac{2}{3}}$

hence correct option is 3

Posted by

shivangi.bhatnagar

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