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Giải thích 2 câu này giúp mik với ạ

Nguyễn Đức Trí · Accepted Answer

Câu 11a) $\overrightarrow{AB}.\overrightarrow{AC}=AB.AC.cos\left(\widehat{\overrightarrow{AB};\overrightarrow{AC}}\right)=1.\sqrt{2}.cos45^o=\sqrt{2}.\dfrac{\sqrt{2}}{2}=1$$\Rightarrow$ Đúngb) $\left(\widehat{\overrightarrow{AB};\overrightarrow{BD}}\right)=90^o+45^o=135^0$$\Rightarrow$ Saic) $\overrightarrow{AC}\perp\overrightarrow{BD}$ (hình vuông) $\Rightarrow\overrightarrow{AC}.\overrightarrow{BD}=0\Rightarrow$ Đúngd) $\left(\overrightarrow{CD}+\overrightarrow{CA}\right)^2=CD^2+CA^2+2CD.CA.cos\left(\widehat{\overrightarrow{CD};\overrightarrow{CA}}\right)$$=1+\left(\sqrt{2}\right)^2+2.1.\sqrt{2}.cos45^o=1+2+2\sqrt{2}.\dfrac{\sqrt{2}}{2}=5\Rightarrow$ ĐúngCâu 12a) $\left(\widehat{\overrightarrow{AB};\overrightarrow{AC}}\right)=\widehat{BAC}=60^o\Rightarrow$ Đúngb) $\overrightarrow{IA}\perp\overrightarrow{BC}$ (tính chất đường trung tuyến tam giác đều)$\Rightarrow\overrightarrow{IA}.\overrightarrow{BC}=0\Rightarrow$ Đúngc) $cos\left(\widehat{\overrightarrow{AB};\overrightarrow{BC}}\right)=cos\left(180^o-\widehat{ABC}\right)=cos120^o=-\dfrac{1}{2}\Rightarrow$ Said) $\left(\overrightarrow{CB}-\overrightarrow{CA}\right)^2=CB^2+CA^2-2CB.CA.cos\left(\widehat{\overrightarrow{CB};\overrightarrow{CA}}\right)$$=2^2+2^2-2.2.2.cos60^o=8-8.\dfrac{1}{2}=4\Rightarrow$ ĐúngCâu trả lời: Câu 11a) $\overrightarrow{AB}.\overrightarrow{AC}=AB.AC.cos\left(\widehat{\overrightarrow{AB};\overrightarrow{AC}}\right)=1.\sqrt{2}.cos45^o=\sqrt{2}.\dfrac{\sqrt{2}}{2}=1$$\Rightarrow$ Đúngb) $\left(\widehat{\overrightarrow{AB};\overrightarrow{BD}}\right)=90^o+45^o=135^0$$\Rightarrow$ Saic) $\overrightarrow{AC}\perp\overrightarrow{BD}$ (hình vuông) $\Rightarrow\overrightarrow{AC}.\overrightarrow{BD}=0\Rightarrow$ Đúngd) $\left(\overrightarrow{CD}+\overrightarrow{CA}\right)^2=CD^2+CA^2+2CD.CA.cos\left(\widehat{\overrightarrow{CD};\overrightarrow{CA}}\right)$$=1+\left(\sqrt{2}\right)^2+2.1.\sqrt{2}.cos45^o=1+2+2\sqrt{2}.\dfrac{\sqrt{2}}{2}=5\Rightarrow$ ĐúngCâu 12a) $\left(\widehat{\overrightarrow{AB};\overrightarrow{AC}}\right)=\widehat{BAC}=60^o\Rightarrow$ Đúngb) $\overrightarrow{IA}\perp\overrightarrow{BC}$ (tính chất đường trung tuyến tam giác đều)$\Rightarrow\overrightarrow{IA}.\overrightarrow{BC}=0\Rightarrow$ Đúngc) $cos\left(\widehat{\overrightarrow{AB};\overrightarrow{BC}}\right)=cos\left(180^o-\widehat{ABC}\right)=cos120^o=-\dfrac{1}{2}\Rightarrow$ Said) $\left(\overrightarrow{CB}-\overrightarrow{CA}\right)^2=CB^2+CA^2-2CB.CA.cos\left(\widehat{\overrightarrow{CB};\overrightarrow{CA}}\right)$$=2^2+2^2-2.2.2.cos60^o=8-8.\dfrac{1}{2}=4\Rightarrow$ Đúng

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