Let
A B C \mathrm{ABC} ABC be a triangle such that
B C → = a → , C A → = b → , A B → = c → , ∣ a → ∣ = 6 2 , ∣ b → ∣ = 2 3 \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} BC = a , CA = b , AB = c , ∣ a ∣ = 6 2 , ∣ b ∣ = 2 3 and
b ⃗ ⋅ c ⃗ = 12 \vec{b} \cdot \vec{c}=12 b ⋅ c = 12 . Consider the statements :
( S 1 ) : ∣ ( a → × b → ) + ( c → × b → ) ∣ − ∣ c ⃗ ∣ = 6 ( 2 2 − 1 ) (\mathrm{S} 1):|(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})+(\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{b}})|-|\vec{c}|=6(2 \sqrt{2}-1) ( S 1 ) : ∣ ( a × b ) + ( c × b ) ∣ − ∣ c ∣ = 6 ( 2 2 − 1 )
( S 2 ) : ∠ A C B = cos − 1 ( 2 3 ) (\mathrm{S} 2): \angle \mathrm{ACB}=\cos ^{-1}\left(\sqrt{\frac{2}{3}}\right) ( S 2 ) : ∠ ACB = cos − 1 ( 3 2 )
Then