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(D) In a triangle $ABC$,$\overrightarrow{BC} + \overrightarrow{CA} + \overrightarrow{AB} = \overrightarrow{0}$,so $\overrightarrow{a} + \overrightarro

Vedclass · Accepted Answer

Both $(S1)$ and $(S2)$ are false.

Vedclass · Answer

(D) In a triangle $ABC$,$\overrightarrow{BC} + \overrightarrow{CA} + \overrightarrow{AB} = \overrightarrow{0}$,so $\overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} = \overrightarrow{0}$.From $\overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} = \overrightarrow{0}$,we have $\overrightarrow{b} + \overrightarrow{c} = -\overrightarrow{a}$.Squaring both sides: $|\overrightarrow{b}|^2 + |\overrightarrow{c}|^2 + 2(\overrightarrow{b} \cdot \overrightarrow{c}) = |\overrightarrow{a}|^2$.Given $|\overrightarrow{a}| = 6\sqrt{2}$,$|\overrightarrow{b}| = 2\sqrt{3}$,and $\overrightarrow{b} \cdot \overrightarrow{c} = 12$,we have $(2\sqrt{3})^2 + |\overrightarrow{c}|^2 + 2(12) = (6\sqrt{2})^2$.$12 + |\overrightarrow{c}|^2 + 24 = 72 \implies |\overrightarrow{c}|^2 = 36 \implies |\overrightarrow{c}| = 6$.For $(S1)$: $|(\overrightarrow{a} 	imes \overrightarrow{b}) + (\overrightarrow{c} 	imes \overrightarrow{b})| - |\overrightarrow{c}| = |(\overrightarrow{a} + \overrightarrow{c}) 	imes \overrightarrow{b}| - |\overrightarrow{c}|$.Since $\overrightarrow{a} + \overrightarrow{c} = -\overrightarrow{b}$,this becomes $|(-\overrightarrow{b}) 	imes \overrightarrow{b}| - |\overrightarrow{c}| = |\overrightarrow{0}| - 6 = -6$.Thus,$(S1)$ is false.For $(S2)$: $\overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} = \overrightarrow{0} \implies \overrightarrow{a} + \overrightarrow{c} = -\overrightarrow{b}$.$|\overrightarrow{a} + \overrightarrow{c}|^2 = |-\overrightarrow{b}|^2 \implies |\overrightarrow{a}|^2 + |\overrightarrow{c}|^2 + 2(\overrightarrow{a} \cdot \overrightarrow{c}) = |\overrightarrow{b}|^2$.$72 + 36 + 2(\overrightarrow{a} \cdot \overrightarrow{c}) = 12 \implies 2(\overrightarrow{a} \cdot \overrightarrow{c}) = -96 \implies \overrightarrow{a} \cdot \overrightarrow{c} = -48$.Using $\cos(\angle ABC) = \frac{\overrightarrow{BA} \cdot \overrightarrow{BC}}{|\overrightarrow{BA}||\overrightarrow{BC}|} = \frac{(-\overrightarrow{c}) \cdot \overrightarrow{a}}{|-\overrightarrow{c}| |\overrightarrow{a}|} = \frac{-(\overrightarrow{a} \cdot \overrightarrow{c})}{6 \cdot 6\sqrt{2}} = \frac{48}{36\sqrt{2}} = \frac{4}{3\sqrt{2}} = \frac{2\sqrt{2}}{3}$.Since $\frac{2\sqrt{2}}{3} 
eq \sqrt{\frac{2}{3}}$,$(S2)$ is false.

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