Question

Let $ABC$ be a triangle such that $\overrightarrow{BC}=\vec{a}$, $\overrightarrow{CA}=\vec{b},\overrightarrow{AB}=\vec{c},|\vec{a}|=6\sqrt{2},|\vec{b}|=2\sqrt{3}$ and $\vec{b}\cdot \vec{c}=12$ Consider the statements : $(S1):|(\vec{a}\times \vec{b})+(\vec{c}\times \vec{b})|-|\vec{c}|=6(2\sqrt{2}-1)$ $(S2):\angle ABC={\cos }^{-1}\left(\sqrt{\frac{2}{3}}\right)$. Then

• A. both (S1) and (S2) are true
• B. only (S1) is true
• C. only (S2) is true
• D. both (S1) and (S2) are false

Answer 1

Answer: (C)

$\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=0$
$\overrightarrow{b}+\overrightarrow{c}=-\overrightarrow{a}$
$|\overrightarrow{b}{|}^2+|\overrightarrow{c}{|}^2+2\overrightarrow{b}\cdot \overrightarrow{c}=|\overrightarrow{a}{|}^2$
$|\overrightarrow{c}{|}^2=36$
$|\overrightarrow{c}{|}^2=6$
$S1:|\vec{a}\times \vec{b}+\overrightarrow{c}\times \overrightarrow{b}|-|\overrightarrow{c}|$
$|(\overrightarrow{a}+\overrightarrow{c})\times \overrightarrow{b}|-|\overrightarrow{c}|$
$|-\overrightarrow{b}\times \overrightarrow{b}|-|\overrightarrow{c}|$
$0-6=-6$
$S2:\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=0$
$\overrightarrow{b}+\overrightarrow{c}=-\overrightarrow{a}$
$|\overrightarrow{a}{|}^2+|\overrightarrow{b}{|}^2-2|\overrightarrow{a}||\overrightarrow{b}|\cos (\angle ACB)=|\overrightarrow{c}{|}^2$
$\cos (\angle ACB)=\sqrt{\frac{2}{3}}$