1.

Do \(\overrightarrow{v}\) cùng phương với \(\overrightarrow{u}\) nên \(\overrightarrow{v}=\left(a;a\right)\) với a là số thực khác 0

Chọn \(M\left(0;0\right)\) là 1 điểm thuộc d

Gọi M' là ảnh của M qua phép tịnh tiến \(\overrightarrow{v}\Rightarrow M'\in d'\)

\(\left\{{}\begin{matrix}x_{M'}=a+0=a\\y_{M'}=a+0=a\end{matrix}\right.\) \(\Rightarrow M'\left(a;a\right)\)

Thay vào pt d' ta được:

\(a+a-4=0\Rightarrow a=2\)

\(\Rightarrow\overrightarrow{v}=\left(2;2\right)\)

\(\Rightarrow\left|\overrightarrow{v}\right|=2\sqrt{2}\)

2.

Gọi \(\overrightarrow{u}=\left(a;b\right)\)

Gọi \(A\left(0;1\right)\) là 1 điểm thuộc d

Gọi A' là ảnh của A qua phép tịnh tiến \(\overrightarrow{u}\Rightarrow A'\in d'\)

Ta có: \(\left\{{}\begin{matrix}x_{A'}=a\\y_{A'}=b+1\end{matrix}\right.\)

Thay tọa độ A' vào pt d' ta được: \(a+b+1-5=0\Leftrightarrow a+b=4\)

Ta có:

\(\left|\overrightarrow{u}\right|=\sqrt{a^2+b^2}\ge\sqrt{\frac{1}{2}\left(a+b\right)^2}=2\sqrt{2}\)

\(\Rightarrow\left|\overrightarrow{u}\right|_{min}=2\sqrt{2}\) khi \(a=b=2\)

3.

Gọi \(\overrightarrow{v}=\left(a;b\right)\Rightarrow a^2+b^2=4\) (1)

Gọi \(A\left(-1;-1\right)\) là 1 điểm thuộc d

Gọi A' là ảnh của A qua phép tịnh tiến \(\overrightarrow{v}\Rightarrow A'\in d'\)

Ta có: \(\left\{{}\begin{matrix}x_{A'}=-1+a\\x_{B'}=-1+b\end{matrix}\right.\)

Thay vào pt d':

\(2\left(a-1\right)+2\left(b-1\right)-1=0\)

\(\Leftrightarrow2a+2b=5\Rightarrow b=\frac{5-2a}{2}\)

Thế vào (1):

\(a^2+\left(\frac{5-2a}{2}\right)^2=4\)

\(\Leftrightarrow8a^2-20a+9=0\)

Pt trên có 2 nghiệm pb nên có 2 vecto thỏa mãn

\(\overrightarrow{x}\) ⊥ \(\overrightarrow{y}\)

⇒ \(\left(\overrightarrow{a}+\overrightarrow{b}\right)\left(\overrightarrow{2a}-\overrightarrow{b}\right)=0\). Đặt \(\left|\overrightarrow{a}\right|=a;\left|\overrightarrow{b}\right|=b\)

⇒ 2a² - \(\overrightarrow{a}.\overrightarrow{b}\) + 2\(\overrightarrow{a}.\overrightarrow{b}\) - b² = 0

⇒ \(\overrightarrow{a}.\overrightarrow{b}\) = b² - 2a² = 4 - 4 = 0

⇒ \(\left(\overrightarrow{a};\overrightarrow{b}\right)=90^0\)

Lời giải:

a)

\(\overrightarrow{x}=\overrightarrow{u}-\overrightarrow{v}=(1-2, 2-2,3-(-1))=(-1,0,4)\)

b)

\(\overrightarrow{x}=\overrightarrow{u}-\overrightarrow{v}+2\overrightarrow{w}=(1-2+2.4,2-2+2.0; 3-(-1)+2(-4))\)

\(=(7, 0, -4)\)

c)

\(\overrightarrow{x}=2\overrightarrow{u}+4\overrightarrow{v}-\overrightarrow{w}=(2.1+4.2-4, 2.2+4.2-0, 2.3+4.(-1)-(-4))\)

\(=(6,12,6)\)

d)

\(2\overrightarrow{x}=3\overrightarrow{u}+\overrightarrow{w}=3(1,2,3)+(4,0,-4)=(3.1+4, 3.2+0,3.3+(-4))\)

\(=(7,6,5)\Rightarrow \overrightarrow{x}=(\frac{7}{2}, 3, \frac{5}{2})\)

e)

\(3\overrightarrow{x}=-2\overrightarrow{u}-\overrightarrow{v}+\overrightarrow{w}=-2(1,2,3)-(2,2,-1)+(4,0,-4)\)

\(=(-2,-4,-6)-(2,2,-1)+(4,0,-4)=(-2-2+4,-4-2+0,-6-(-1)+(-4))\)

\(=(0,-6,-9)\Rightarrow \overrightarrow{x}=(0,-2,-3)\)

\(\overrightarrow{a}\perp\overrightarrow{b}\Rightarrow\overrightarrow{a}.\overrightarrow{b}=0\)

\(\left(2\overrightarrow{a}-\overrightarrow{b}\right)\left(\overrightarrow{a}+\overrightarrow{b}\right)=2a^2+2\overrightarrow{a}.\overrightarrow{b}-\overrightarrow{a}.\overrightarrow{b}-b^2\)

\(=2a^2-b^2+\overrightarrow{a}.\overrightarrow{b}\)

\(=2.1-2+0=0\)

\(\Rightarrow\left(2\overrightarrow{a}-\overrightarrow{b}\right)\perp\left(\overrightarrow{a}+\overrightarrow{b}\right)\)

1. \(\Leftrightarrow\left\{{}\begin{matrix}\frac{1}{x+1}=1\\\frac{1}{y-1}=\frac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=3\end{matrix}\right.\)

2. \(\Leftrightarrow\overrightarrow{a}\cdot\overrightarrow{b}=0\)

\(\Leftrightarrow4m+\left(-2\right)\cdot\left(-1\right)=0\)

\(\Leftrightarrow m=-\frac{1}{2}\)

1. a) D = [1;4] \{2;3}

b) D = (0;+∞)

2.

\(2\overrightarrow{a}\)= (2;4) và \(3\overrightarrow{b}\) = (9;12)

⇒ \(2\overrightarrow{a}\) + \(3\overrightarrow{b}\) = (2+9; 4+12)

⇔ (11; 16)

Vậy \(\overrightarrow{m}\) = (11;16)

1/ \(\overrightarrow{AB}^2-\overrightarrow{AD}^2=\overrightarrow{BC}^2-\overrightarrow{CD}^2\)

\(\Leftrightarrow\left(\overrightarrow{AB}+\overrightarrow{AD}\right)\left(\overrightarrow{AB}-\overrightarrow{AD}\right)=\left(\overrightarrow{BC}+\overrightarrow{CD}\right)\left(\overrightarrow{BC}-\overrightarrow{CD}\right)\)

\(\Leftrightarrow\left(\overrightarrow{AB}+\overrightarrow{AD}\right).\overrightarrow{DB}=\overrightarrow{BD}\left(\overrightarrow{BC}-\overrightarrow{CD}\right)=\overrightarrow{DB}\left(\overrightarrow{CB}+\overrightarrow{CD}\right)\)

Gọi M là trung điểm BD

\(\Rightarrow2\overrightarrow{AM}.\overrightarrow{DB}=2\overrightarrow{CM}.\overrightarrow{DB}\)

\(\Leftrightarrow\overrightarrow{DB}.\left(\overrightarrow{AM}-\overrightarrow{CM}\right)=0\)

\(\Leftrightarrow\overrightarrow{BD}.\overrightarrow{AC}=0\)

2/ \(A=\left|\overrightarrow{a}-\overrightarrow{b}\right|\Rightarrow A^2=\overrightarrow{a}^2-2\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}^2\)

\(=a^2+b^2-2ab.cos\left(\overrightarrow{a};\overrightarrow{b}\right)=4^2+5^2-2.4.5.cos120^0=61\)

\(\Rightarrow A=\sqrt{61}\)

b/ \(B=\left|2\overrightarrow{a}+\overrightarrow{b}\right|\Rightarrow B^2=4a^2+b^2+4\overrightarrow{a}.\overrightarrow{b}\)

\(=4a^2+b^2+4ab.cos120^0=49\)

\(\Rightarrow B=7\)

3/ \(\left|\overrightarrow{x}\right|=\left|\overrightarrow{a}-2\overrightarrow{b}\right|\Rightarrow\left|\overrightarrow{x}\right|^2=a^2+4b^2-4\overrightarrow{a}.\overrightarrow{b}=12\)

\(\Rightarrow\left|\overrightarrow{x}\right|=2\sqrt{3}\)

\(\left|\overrightarrow{y}\right|^2=a^2+b^2-2\overrightarrow{a}.\overrightarrow{b}=5\Rightarrow\left|\overrightarrow{y}\right|=\sqrt{5}\)

\(\overrightarrow{x}.\overrightarrow{y}=\left(\overrightarrow{a}-2\overrightarrow{b}\right)\left(\overrightarrow{a}-\overrightarrow{b}\right)=a^2+2b^2-3\overrightarrow{a}.\overrightarrow{b}=4\)

\(\Rightarrow cos\alpha=\frac{\overrightarrow{x}.\overrightarrow{y}}{\left|\overrightarrow{x}\right|.\left|\overrightarrow{y}\right|}=\frac{4}{2\sqrt{15}}=\frac{2\sqrt{15}}{15}\)

(1); vecto u=2*vecto a-vecto b

=>\(\left\{{}\begin{matrix}x=2\cdot1-0=2\\y=2\cdot\left(-4\right)-2=-10\end{matrix}\right.\)

(2): vecto u=-2*vecto a+vecto b

=>\(\left\{{}\begin{matrix}x=-2\cdot\left(-7\right)+4=18\\y=-2\cdot3+1=-5\end{matrix}\right.\)

(3): vecto a=2*vecto u-5*vecto v

\(\Leftrightarrow\left\{{}\begin{matrix}a=2\cdot\left(-5\right)-5\cdot0=-10\\b=2\cdot4-5\cdot\left(-3\right)=15+8=23\end{matrix}\right.\)

(4): vecto OM=(x;y)

2 vecto OA-5 vecto OB=(-18;37)

=>x=-18; y=37

=>x+y=19

Đường tròn \(\left(C\right)\) tâm \(A\left(8;3\right)\) bán kính \(R=\sqrt{10}\)

Gọi B là tâm của (C') thì B là ảnh của A qua phép tịnh tiến \(\overrightarrow{v}\)

\(\Rightarrow B\left(13;10\right)\)

Phương trình (C'):

\(\left(x-13\right)^2+\left(x-10\right)^2=10\)

\(\overrightarrow{a}+\overrightarrow{b}+3\overrightarrow{c}=\overrightarrow{0}\Leftrightarrow\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=-2\overrightarrow{c}\)

\(\Leftrightarrow\left(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\right)^2=\left(-2\overrightarrow{c}\right)^2\)

\(\Leftrightarrow\overrightarrow{a}^2+\overrightarrow{b}^2+\overrightarrow{c}^2+2\left(\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{a}\right)=4\overrightarrow{c}^2\)

\(\Leftrightarrow A=\dfrac{4x^2-\left(x^2+y^2+z^2\right)}{2}=\dfrac{3x^2-y^2-z^2}{2}\)