Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1)\(VT=\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OD}=\overrightarrow{CO}+\overrightarrow{DO}+\overrightarrow{OC}+\overrightarrow{OC}=\overrightarrow{CO}+\overrightarrow{OC}+\overrightarrow{DO}+\overrightarrow{OD}=\overrightarrow{0}\)
2)\(VT=\overrightarrow{DA}-\overrightarrow{DB}+\overrightarrow{DC}=\overrightarrow{BA}+\overrightarrow{DC}=\overrightarrow{0}\)
3)\(VT=\overrightarrow{DO}+\overrightarrow{AO}=\overrightarrow{OB}+\overrightarrow{AO}=\overrightarrow{AB}\)
4)\(\overrightarrow{MA}+\overrightarrow{MC}=\overrightarrow{MB}+\overrightarrow{BA}+\overrightarrow{MD}+\overrightarrow{DC}=\overrightarrow{MB}+\overrightarrow{MD}\left(đpcm\right)=\overrightarrow{MO}+\overrightarrow{OB}+\overrightarrow{MO}+\overrightarrow{OD}=2\overrightarrow{MO}\left(đpcm\right)\)
Chúc bạn học tốt!!!!!
Đăng kí kênh Youtube 'Ban Mai Anime' giúp mình nhé!!!!
\(\overrightarrow{AB}+\overrightarrow{CD}=\overrightarrow{AB}+\overrightarrow{CB}+\overrightarrow{BD}=\overrightarrow{AB}+\overrightarrow{BD}+\overrightarrow{CB}=\overrightarrow{AD}+\overrightarrow{CB}\)
\(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OD}=\left(\overrightarrow{OE}+\overrightarrow{EA}\right)+\left(\overrightarrow{OF}+\overrightarrow{FB}\right)+\left(\overrightarrow{OE}+\overrightarrow{EC}\right)+\left(\overrightarrow{OF}+\overrightarrow{FD}\right)\)
\(=2\left(\overrightarrow{OE}+\overrightarrow{EF}\right)+\left(\overrightarrow{EA}+\overrightarrow{EC}\right)+\left(\overrightarrow{FB}+\overrightarrow{FD}\right)\)
\(=2.\overrightarrow{0}+\overrightarrow{0}+\overrightarrow{0}=\overrightarrow{0}\)
a) \(AC = BD = \sqrt {A{B^2} + A{D^2}} \\= \sqrt {{{\left( {2a} \right)}^2} + {a^2}} = a\sqrt 5 \)
\(\cos \left( {\overrightarrow {AB} ,\overrightarrow {AO} } \right) = \cos \widehat {OAB} =\\ \cos \widehat {CAB} = \frac{{AB}}{{AC}} = \frac{{2a}}{{a\sqrt 5 }} = \frac{{2\sqrt 5 }}{5}\)
\(\begin{array}{l}\overrightarrow {AB} .\overrightarrow {AO} = \left| {\overrightarrow {AB} } \right|.\left| {\overrightarrow {AO} } \right|.\cos \left( {\overrightarrow {AB} ,\overrightarrow {AO} } \right) \\= AB.\frac{1}{2}AC.\cos \left( {\overrightarrow {AB} ,\overrightarrow {AO} } \right)\\ = 2a.\frac{1}{2}.a\sqrt 5 .\frac{{2\sqrt 5 }}{5} = 2{a^2}\end{array}\)
b) \(AB \bot AD \Rightarrow \left( {\overrightarrow {AB} ,\overrightarrow {AD} } \right) = 90^o \Rightarrow \cos \left( {\overrightarrow {AB} ,\overrightarrow {AD} } \right) =0 \Rightarrow \overrightarrow {AB} .\overrightarrow {AD} = 0\)
1: \(=\left|\overrightarrow{CO}-\overrightarrow{CB}\right|=BO=\dfrac{a\sqrt{2}}{2}\)
Đề là \(AB=4\) hay \(AD=4\) nhỉ? Sao lại có 2 kích thước của AD?