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a, \(\left(\overrightarrow{AC}-\overrightarrow{AB}\right)^2=\overrightarrow{BC}^2\)
\(\Leftrightarrow AC^2+AB^2-2\overrightarrow{AB}.\overrightarrow{AC}=BC^2\)
\(\Leftrightarrow2\overrightarrow{AB}.\overrightarrow{AC}=AB^2+AC^2-BC^2\)
\(\Rightarrow\overrightarrow{AB}.\overrightarrow{AC}=\dfrac{AB^2+AC^2-BC^2}{2}=\dfrac{5^2+8^2-7^2}{2}=20\)
b, \(2\overrightarrow{CA}.\overrightarrow{CB}=CA^2+CB^2-BC^2=CA^2\)
\(\Rightarrow\overrightarrow{CA}.\overrightarrow{CB}=\dfrac{CA^2}{2}=\dfrac{8^2}{2}=32\)
Lời giải:
a)
\(\overrightarrow{AC}-\overrightarrow{AB}=\overrightarrow{BC}\)
\(\Rightarrow (\overrightarrow{AC}-\overrightarrow{AB})^2=\overrightarrow{BC}^2\Leftrightarrow AB^2+AC^2-2\overrightarrow{AC}.\overrightarrow{AB}=BC^2\)
\(\Leftrightarrow 2\overrightarrow{AB}.\overrightarrow{AC}=AB^2+AC^2-BC^2\) (đpcm)
Ta có:
\(\overrightarrow{AB}.\overrightarrow{AC}=\frac{AB^2+AC^2-BC^2}{2}=\frac{5^2+8^2-7^2}{2}=20\)
\(\cos \angle A=\frac{\overrightarrow{AB}.\overrightarrow{AC}}{|\overrightarrow{AB}|.|\overrightarrow{AC}|}=\frac{20}{5.8}=\frac{1}{2}\)
\(\Rightarrow \angle A=60^0\)
b)
Tương tự phần a, \(\overrightarrow{CA}.\overrightarrow{CB}=\frac{CA^2+CB^2-AB^2}{2}=\frac{8^2+7^2-5^2}{2}=44\)
1.
\(\overrightarrow{AB}.\overrightarrow{BC}=\overrightarrow{AB}.\left(\overrightarrow{BA}+\overrightarrow{AC}\right)=\overrightarrow{AB}.\left(-\overrightarrow{AB}\right)+\overrightarrow{AB}.\overrightarrow{AC}=-AB^2=-25\)
2.
\(\overrightarrow{AB}.\overrightarrow{BD}=\overrightarrow{AB}\left(\overrightarrow{BA}+\overrightarrow{AD}\right)=-\overrightarrow{AB}.\overrightarrow{AB}+\overrightarrow{AB}.\overrightarrow{AD}=-AB^2+0=-64\)
a) \(\overrightarrow {AB} .\overrightarrow {AC} = 2.3.\cos \widehat {BAC} = 6.\cos {60^o} = 3\)
b)
Ta có: \(\overrightarrow {AB} + \overrightarrow {AC} = 2\overrightarrow {AM} \)(do M là trung điểm của BC)
\( \Leftrightarrow \overrightarrow {AM} = \frac{1}{2}\overrightarrow {AB} + \frac{1}{2}\overrightarrow {AC} \)
+) \(\overrightarrow {BD} = \overrightarrow {AD} - \overrightarrow {AB} = \frac{7}{{12}}\overrightarrow {AC} - \overrightarrow {AB} \)
c) Ta có:
\(\begin{array}{l}\overrightarrow {AM} .\overrightarrow {BD} = \left( {\frac{1}{2}\overrightarrow {AB} + \frac{1}{2}\overrightarrow {AC} } \right)\left( {\frac{7}{{12}}\overrightarrow {AC} - \overrightarrow {AB} } \right)\\ = \frac{7}{{24}}\overrightarrow {AB} .\overrightarrow {AC} - \frac{1}{2}{\overrightarrow {AB} ^2} + \frac{7}{{24}}{\overrightarrow {AC} ^2} - \frac{1}{2}\overrightarrow {AC} .\overrightarrow {AB} \\ = - \frac{1}{2}A{B^2} + \frac{7}{{24}}A{C^2} - \frac{5}{{24}}\overrightarrow {AB} .\overrightarrow {AC} \\ = - \frac{1}{2}{.2^2} + \frac{7}{{24}}{.3^2} - \frac{5}{{24}}.3\\ = 0\end{array}\)
\( \Rightarrow AM \bot BD\)
\(cosA=\dfrac{AB^2+AC^2-BC^2}{2AB.AC}=\dfrac{5}{24}\)
\(\Rightarrow\overrightarrow{AB}.\overrightarrow{AC}=AB.AC.cosA=10a^2\)
+) Ta có: \(AB \bot AC \Rightarrow \overrightarrow {AB} \bot \overrightarrow {AC} \Rightarrow \overrightarrow {AB} .\overrightarrow {AC} = 0\)
+) \(\overrightarrow {AC} .\overrightarrow {BC} = \left| {\overrightarrow {AC} } \right|.\left| {\overline {BC} } \right|.\cos \left( {\overrightarrow {AC} ,\overrightarrow {BC} } \right)\)
Ta có: \(BC = \sqrt {A{B^2} + A{C^2}} = \sqrt 2 \Leftrightarrow \sqrt {2A{C^2}} = \sqrt 2 \)\( \Rightarrow AC = 1\)
\( \Rightarrow \overrightarrow {AC} .\overrightarrow {BC} = 1.\sqrt 2 .\cos \left( {45^\circ } \right) = 1\)
+) \(\overrightarrow {BA} .\overrightarrow {BC} = \left| {\overrightarrow {BA} } \right|.\left| {\overrightarrow {BC} } \right|.\cos \left( {\overrightarrow {BA} ,\overrightarrow {BC} } \right) = 1.\sqrt 2 .\cos \left( {45^\circ } \right) = 1\)
Tham khảo:
a) \(\)\(\overrightarrow {BA} + \overrightarrow {AC} = \overrightarrow {BC} \Rightarrow \left| {\overrightarrow {BC} } \right| = BC = a\)
b) Dựng hình bình hành ABDC, giao điểm của hai đường chéo là O ta có:
\(\overrightarrow {AB} + \overrightarrow {AC} = \overrightarrow {AD} \)
\(AD = 2AO = 2\sqrt {A{B^2} - B{O^2}} = 2\sqrt {{a^2} - {{\left( {\frac{a}{2}} \right)}^2}} = a\sqrt 3 \)
\( \Rightarrow \left| {\overrightarrow {AB} + \overrightarrow {AC} } \right| = \left| {\overrightarrow {AD} } \right| = AD = a\sqrt 3 \)
c) \(\overrightarrow {BA} - \overrightarrow {BC} = \overrightarrow {BA} + \overrightarrow {CB} = \overrightarrow {CB} + \overrightarrow {BA} = \overrightarrow {CA} \)
\( \Rightarrow \left| {\overrightarrow {BA} - \overrightarrow {BC} } \right| = \left| {\overrightarrow {CA} } \right| = CA = a\)