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b) \(S=\frac{1}{2}\sqrt{AB^2.AC^2-\left(\overrightarrow{AB}.\overrightarrow{AC}\right)^2}\)
\(=\frac{1}{2}\sqrt{AB^2.AC^2-AB^2.AC^2.cos^2A}\)
\(=\frac{1}{2}\sqrt{AB^2AC^2.sin^2A}\)
\(=\frac{1}{2}.AB.AC.\sin A\) (đpcm)
a) \(\overrightarrow{u}=3\overrightarrow{a}+2\overrightarrow{b}-4\overrightarrow{c}=3\left(2;1\right)+2\left(3;-4\right)-4\left(-7;2\right)\)
\(=\left(6;3\right)+\left(6;-8\right)-\left(-28;8\right)\)
\(=\left(6+6+28;3-8-8\right)=\left(40;-13\right)\).
b) \(\overrightarrow{x}+\overrightarrow{a}=\overrightarrow{b}-\overrightarrow{c}\Leftrightarrow\overrightarrow{x}=\overrightarrow{b}-\overrightarrow{c}-\overrightarrow{a}\)
\(\Leftrightarrow\overrightarrow{x}=\left(3;-4\right)-\left(-7;2\right)-\left(2;1\right)\)
\(\Leftrightarrow\overrightarrow{x}=\left(3+7-2;-4-2-1\right)\)
\(\Leftrightarrow\overrightarrow{x}=\left(8;-7\right)\).
c) Có \(\overrightarrow{c}\left(-7;2\right)=k\overrightarrow{a}+h\overrightarrow{b}=k\left(2;1\right)+h\left(3;-4\right)\)
\(=\left(2k+3h;k-4h\right)\).
Từ đó suy ra: \(\left\{{}\begin{matrix}2k+3h=-7\\k-4h=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}k=-2\\h=-1\end{matrix}\right.\).
a) \(\overrightarrow{a}=2\overrightarrow{u}+3\overrightarrow{v}=2\left(3;-4\right)+3\left(2;5\right)=\left(6;-8\right)+\left(6;15\right)\)\(=\left(12;7\right)\).
b) \(\overrightarrow{b}=\overrightarrow{u}-\overrightarrow{v}=\left(3;-4\right)-\left(2;5\right)=\left(1;-9\right)\).
c) Hai véc tơ \(\overrightarrow{c}=\left(m;10\right)\) và \(\overrightarrow{v}\) cùng phương khi và chỉ khi:
\(\dfrac{m}{2}=\dfrac{10}{5}=2\Rightarrow m=4\).
\(\overrightarrow{x}=\overrightarrow{a}+\overrightarrow{b}=\left(1+0;-2+3\right)=\left(1;1\right)\).
\(\overrightarrow{y}=\overrightarrow{a}-\overrightarrow{b}=\left(0-1;3-\left(-2\right)\right)=\left(-1;5\right)\).
\(\overrightarrow{z}=3\overrightarrow{a}-4\overrightarrow{b}=3\left(1;-2\right)-4\left(0;3\right)=\left(3;-6\right)-\left(0;12\right)\)\(=\left(3;-18\right)\).
Hok nhanh phết, chưa j đã đến phần toạ độ vecto r
1/ \(\overrightarrow{MB}=\left(x_B-x_M;y_B-y_M\right)=\left(2-x_M;3-y_M\right)\)
\(\Rightarrow2\overrightarrow{MB}=\left(4-2x_M;6-2y_M\right)\)
\(\overrightarrow{3MC}=\left(3x_C-3x_M;3y_C-3y_M\right)=\left(-3-3x_M;6-3y_M\right)\)
\(\Rightarrow2\overrightarrow{MB}+3\overrightarrow{MC}=\left(4-2x_M-3-3x_M;6-2y_M+6-3y_M\right)=0\)
\(\Leftrightarrow\left(1-5x_M;12-5y_M\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}1-5x_M=0\\12-5y_M=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_M=\frac{1}{5}\\y_M=\frac{12}{5}\end{matrix}\right.\Rightarrow M\left(\frac{1}{5};\frac{12}{5}\right)\)
2/ \(\overrightarrow{m}=2\left(1;2\right)+3\left(3;4\right)=\left(2+9;4+12\right)=\left(11;16\right)\)
3/ \(\overrightarrow{AB}=\left(x_B-x_A;y_B-y_A\right)=\left(-5-3;4+2\right)=\left(-8;6\right)\)
\(\overrightarrow{AC}=\left(x_C-x_A;y_C-y_A\right)=\left(\frac{1}{3}-3;0+2\right)=\left(-\frac{8}{3};2\right)\)
\(\Rightarrow x=\frac{\overrightarrow{AB}}{\overrightarrow{AC}}=\frac{\left(-8;6\right)}{\left(-\frac{8}{3};2\right)}=3\)
Câu 4 tương tự
Câu 5 vt lại đề bài nhé bn, nghe nó vô lý sao á, m,n ở đâu ra vậy, cả A,B,C nx
\(\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=\overrightarrow{0}\Rightarrow\left(\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}\right)^2=0\)
\(\Rightarrow-2\left(\overrightarrow{GA}.\overrightarrow{GB}+\overrightarrow{GB}.\overrightarrow{GC}+\overrightarrow{GC}.\overrightarrow{GA}\right)=GA^2+GB^2+GC^2\)
\(\Rightarrow\overrightarrow{GA}.\overrightarrow{GB}+\overrightarrow{GB}.\overrightarrow{GC}+\overrightarrow{GC}.\overrightarrow{GA}=-\frac{1}{2}\left(\frac{2}{3}m_a^2+\frac{2}{3}m_b^2+\frac{2}{3}m_c^2\right)\)
\(=-\frac{1}{6}\left(AB^2+BC^2+CA^2\right)\)
Hình như đề bài sai dấu?
a: \(\Leftrightarrow\left\{{}\begin{matrix}x+3y=5\\2x-y=6\end{matrix}\right.\)=>x=23/7; y=4/7
b: \(2\cdot\overrightarrow{A}+3\cdot\overrightarrow{B}\)
\(=\left(2\cdot1+3\cdot3;2\cdot2+3\cdot\left(-1\right)\right)\)
=(11;1)
c: \(\overrightarrow{A}\cdot\overrightarrow{B}=\left(3;-2\right)\)