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Ta có:
\(\overrightarrow{AM}=\dfrac{1}{2}\left(\overrightarrow{AO}+\overrightarrow{AB}\right)\)
\(\Leftrightarrow\overrightarrow{AM}=\dfrac{1}{2}\left(\overrightarrow{AO}+\overrightarrow{AO}+\overrightarrow{OB}\right)\)
\(\Leftrightarrow\overrightarrow{AM}=\dfrac{1}{2}\left(2\overrightarrow{AO}+\overrightarrow{OB}\right)\)
\(\Leftrightarrow\overrightarrow{AM}=\overrightarrow{AO}+\dfrac{1}{2}\overrightarrow{OB}\)
\(\Leftrightarrow\overrightarrow{AM}=\dfrac{1}{2}\overrightarrow{OB}-\overrightarrow{OA}\)
\(\RightarrowĐPCM\)
Câu b ) Bạn làm tương tự câu a , ta có vecto BN = 1/2 (BO +BC ) , rồi là như câu a
chúc bạn hok tốt
1.
Gọi G là trọng tâm tam giác
\(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}=\overrightarrow{0}\)
\(\Leftrightarrow3\overrightarrow{OG}=\overrightarrow{0}\)
\(\Leftrightarrow O\equiv G\)
\(\Rightarrow O\) là trọng tâm tam giác ABC
\(\Rightarrow\Delta ABC\) đều
Gọi độ dài các cạnh tam giác là a
\(\overrightarrow{BN}.\overrightarrow{AM}=\dfrac{1}{4}\left(\overrightarrow{AB}+\overrightarrow{AC}\right)\left(\overrightarrow{BA}+\overrightarrow{BC}\right)=-\dfrac{1}{4}a^2-\dfrac{1}{8}a^2-\dfrac{1}{8}a^2+\dfrac{1}{2}a^2=0\)
Mặt khác \(\overrightarrow{BN}.\overrightarrow{AM}=BN.AM.cos\left(\overrightarrow{AM};\overrightarrow{BN}\right)\)
\(\Rightarrow BN.AM.cos\left(\overrightarrow{AM};\overrightarrow{BN}\right)=0\Rightarrow cos\left(\overrightarrow{AM};\overrightarrow{BN}\right)=0\Rightarrow\left(\overrightarrow{AM};\overrightarrow{BN}\right)=90^o\)
\(BD=\dfrac{AB}{cos45^o}=\dfrac{a}{\dfrac{\sqrt{2}}{2}}=a\sqrt{2}\)
\(\overrightarrow{BQ}.\overrightarrow{BP}=\dfrac{1}{4}\left(\overrightarrow{BA}+\overrightarrow{BD}\right)\left(\overrightarrow{BC}+\overrightarrow{BD}\right)\)
\(=\dfrac{1}{4}BA.BC.cos90^o+\dfrac{1}{4}BA.BD.cos45^o+\dfrac{1}{4}BD.BC.cos45^o+\dfrac{1}{4}BD^2\)
\(=\dfrac{1}{4}a^2+\dfrac{1}{4}a^2+\dfrac{1}{2}a^2=a^2\)
a: \(\overrightarrow{AM}+\overrightarrow{BN}=\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{BC}=\dfrac{1}{2}\overrightarrow{AC}\)
b: \(=\dfrac{1}{2}\overrightarrow{AC}+\dfrac{1}{2}\overrightarrow{AC}+\dfrac{1}{2}\overrightarrow{BA}\)
\(=\overrightarrow{AC}+\dfrac{1}{2}\overrightarrow{BA}\)
c: \(\overrightarrow{AM}+\overrightarrow{BN}+\overrightarrow{CP}\)
\(=\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{BC}+\dfrac{1}{2}\overrightarrow{CA}\)
\(=\dfrac{1}{2}\left(\overrightarrow{AC}+\overrightarrow{CA}\right)=\overrightarrow{0}\)
a.
Do M là trung điểm OB \(\Rightarrow\overrightarrow{OM}=\dfrac{1}{2}\overrightarrow{OB}\)
\(\Rightarrow\overrightarrow{AM}=\overrightarrow{AO}+\overrightarrow{OM}=-\overrightarrow{OA}+\dfrac{1}{2}\overrightarrow{OB}\)
b.
Do N là trung điểm OC \(\Rightarrow\overrightarrow{ON}=\dfrac{1}{2}\overrightarrow{OC}\)
\(\Rightarrow\overrightarrow{BN}=\overrightarrow{BO}+\overrightarrow{ON}=-\overrightarrow{OB}+\dfrac{1}{2}\overrightarrow{OC}\)
\(\overrightarrow{MN}=\overrightarrow{MO}+\overrightarrow{ON}=-\overrightarrow{OM}+\overrightarrow{ON}=-\dfrac{1}{2}\overrightarrow{OB}+\dfrac{1}{2}\overrightarrow{OC}\)
Tham khảo:
Dễ thấy: \(\overrightarrow {OA} = \overrightarrow {OM} + \overrightarrow {MA} \); \(\overrightarrow {OB} = \overrightarrow {OM} + \overrightarrow {MB} \)
Tương tự: \(\overrightarrow {OC} = \overrightarrow {ON} + \overrightarrow {NC} \); \(\overrightarrow {OD} = \overrightarrow {ON} + \overrightarrow {ND} \)
\(\begin{array}{l} \Rightarrow \overrightarrow {OA} + \overrightarrow {OB} + \overrightarrow {OC} + \overrightarrow {OD} = \left( {\overrightarrow {OM} + \overrightarrow {MA} } \right) + \left( {\overrightarrow {OM} + \overrightarrow {MB} } \right) + \left( {\overrightarrow {ON} + \overrightarrow {NC} } \right) + \left( {\overrightarrow {ON} + \overrightarrow {ND} } \right)\\ = \left( {\overrightarrow {OM} + \overrightarrow {OM} + \overrightarrow {MA} + \overrightarrow {MB} } \right) + \left( {\overrightarrow {ON} + \overrightarrow {ON} + \overrightarrow {NC} + \overrightarrow {ND} } \right)\\ = \overrightarrow {OM} + \overrightarrow {OM} + \overrightarrow {ON} + \overrightarrow {ON} \\ = \left( {\overrightarrow {OM} + \overrightarrow {ON} } \right) + \left( {\overrightarrow {OM} + \overrightarrow {ON} } \right)\\ = \overrightarrow 0 + \overrightarrow 0 \\ = \overrightarrow 0 .\end{array}\)
\(3\overrightarrow{AC}=2\overrightarrow{CB}\leftrightarrow3\overrightarrow{AO}+3\overrightarrow{OC}=2\overrightarrow{CO}+2\overrightarrow{OB}\)
\(\rightarrow5\overrightarrow{OC}=3\overrightarrow{OA}+2\overrightarrow{OB}\)
\(\rightarrow\overrightarrow{OC}=\dfrac{3}{5}\overrightarrow{OA}+\dfrac{2}{5}\overrightarrow{OB}\)
câu 2 ( các kí hiệu vecto khi lm bài thỳ b tự viết nhé mk k viết kí hiệu để trả lời cho nhanh hỳ hỳ )
OA+ OB + OC = OA'+ OB' + OC'
<=> OA - OA' + OB - OB' + OC - OC' = 0
<=> A'A + B'B + C'C = 0
<=> 2 ( BA + CB + AC ) = 0
<=> 2 ( CB + BA + AC ) = 0
<=> 2 ( CA + AC ) = 0
<=> 0 = 0 ( luôn đúng )
câu 1 ( các kí hiệu vecto b cx tự viết nhá )
VT = OD + OC = OA + AD + OB + BC = OA + OB + AD + BC = BO + OB + AD + BC = 0 + AD + BC = AD + BC = VP ( đpcm)
\(\overrightarrow{AB}+\overrightarrow{CD}=\overrightarrow{AB}+\overrightarrow{CB}+\overrightarrow{BD}=\overrightarrow{AB}+\overrightarrow{BD}+\overrightarrow{CB}=\overrightarrow{AD}+\overrightarrow{CB}\)
\(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OD}=\left(\overrightarrow{OE}+\overrightarrow{EA}\right)+\left(\overrightarrow{OF}+\overrightarrow{FB}\right)+\left(\overrightarrow{OE}+\overrightarrow{EC}\right)+\left(\overrightarrow{OF}+\overrightarrow{FD}\right)\)
\(=2\left(\overrightarrow{OE}+\overrightarrow{EF}\right)+\left(\overrightarrow{EA}+\overrightarrow{EC}\right)+\left(\overrightarrow{FB}+\overrightarrow{FD}\right)\)
\(=2.\overrightarrow{0}+\overrightarrow{0}+\overrightarrow{0}=\overrightarrow{0}\)