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a: A(1;3); B(-2;5); C(-4;0)
\(\overrightarrow{AB}=\left(-2-1;5-3\right)=\left(-3;2\right)\)
\(\overrightarrow{AC}=\left(-4-1;0-3\right)=\left(-5;-3\right)\)
\(\overrightarrow{BC}=\left(-4+2;0-5\right)=\left(-2;-5\right)\)
\(\overrightarrow{CB}=\left(-2+4;5-0\right)=\left(2;5\right)\)
b: \(\overrightarrow{AB}\cdot\overrightarrow{CB}=-3\cdot2+2\cdot5=-6+10=4\)
\(\overrightarrow{AC}\cdot\overrightarrow{BC}=\left(-5\right)\cdot\left(-2\right)+\left(-3\right)\cdot\left(-5\right)=10+15=25\)
c: \(\overrightarrow{AB}=\left(-3;2\right)\)
=>\(AB=\sqrt{\left(-3\right)^2+2^2}=\sqrt{13}\)
\(\overrightarrow{BC}=\left(-2;-5\right)\)
=>\(BC=\sqrt{\left(-2\right)^2+\left(-5\right)^2}=\sqrt{4+25}=\sqrt{29}\)
e: \(\overrightarrow{AB}+2\cdot\overrightarrow{CB}\) =\(\left(-3+2\cdot2;2+2\cdot5\right)\)
=(-3+4;2+10)
=(1;12)
\(\overrightarrow{a}.\overrightarrow{b}=\left|\overrightarrow{a}\right|.\left|\overrightarrow{b}\right|.cos\left(\overrightarrow{a};\overrightarrow{b}\right)\)
a/ \(\overrightarrow{a}.\overrightarrow{b}=8.\sqrt{3}.cos30^0=12\)
b/ \(\overrightarrow{a}.\overrightarrow{b}=\sqrt{2}.6.cos45^0=6\)
c/ \(\overrightarrow{a}.\overrightarrow{b}=9.10.cos60^0=45\)
d/ \(\overrightarrow{a}.\overrightarrow{b}=5.6.cos120^0=-15\)
Gọi \(M\left(x;y\right)\Rightarrow\left\{{}\begin{matrix}\overrightarrow{MA}=\left(1-x;3-y\right)\\\overrightarrow{MB}=\left(4-x;-y\right)\\\overrightarrow{MC}=\left(2-x;-5-y\right)\end{matrix}\right.\)
\(\Rightarrow\overrightarrow{MA}+\overrightarrow{MB}-3\overrightarrow{MC}=\left(x-1;y+18\right)\)
\(\Rightarrow\left\{{}\begin{matrix}x-1=0\\y+18=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=1\\y=-18\end{matrix}\right.\)
\(\Rightarrow M\left(1;-18\right)\)
Đề bài sai bạn, \(\overrightarrow{a}\perp\overrightarrow{d}\) thì \(\overrightarrow{a}.\overrightarrow{d}=0\) chứ làm gì có chuyện \(\overrightarrow{a}.\overrightarrow{d}=20\) nữa