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

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

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

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

vecto x=vecto AB+vecto AC-vecto BC

=vecto AB+vecto AC+vecto CB

=vecto AB+vecto AB

=2*vecto AB

=>|vecto x|=2*3a=6a

A. Ta có: \(\overrightarrow u .\overrightarrow v  = 2.4 + 3.6 = 26 \ne 0\) nên \(\overrightarrow u \) và \(\overrightarrow v \) không vuông góc với nhau.

B.  Ta có: \(\overrightarrow a .\overrightarrow b  = 1.( - 1) + ( - 1).1 =  - 2 \ne 0\) nên \(\overrightarrow a \) và \(\overrightarrow b \) không vuông góc với nhau.

C. Ta có: \(\overrightarrow z .\overrightarrow t  = a.( - b) + b.a = 0\) nên \(\overrightarrow z \) và \(\overrightarrow t \) vuông góc với nhau.

Chọn đáp án C

D. Ta có: \(\overrightarrow n .\overrightarrow k  = 1.2 + 1.0 = 2 \ne 0\) nên \(\overrightarrow n \) và \(\overrightarrow k \) không vuông góc với nhau.

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

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

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

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

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

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

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

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

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

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

\(\overrightarrow{u}\overrightarrow{v}=0\Rightarrow\left(\overrightarrow{a}+3\overrightarrow{b}\right)\left(7\overrightarrow{a}-5\overrightarrow{b}\right)=7a^2+16\overrightarrow{a}\overrightarrow{b}-15b^2=0\left(1\right)\)

\(\overrightarrow{x}\overrightarrow{y}=0\Rightarrow\left(\overrightarrow{a}-4\overrightarrow{b}\right)\left(7\overrightarrow{a}-2\overrightarrow{b}\right)=7a^2-30\overrightarrow{a}\overrightarrow{b}+8b^2=0\left(2\right)\)

(1) và (2): \(\left\{{}\begin{matrix}7a^2+16\overrightarrow{a}\overrightarrow{b}-15b^2=0\\7a^2-30\overrightarrow{a}\overrightarrow{b}+8b^2=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\overrightarrow{a}\overrightarrow{b}=\frac{b^2}{2}\\a^2=b^2\Rightarrow\left|a\right|=\left|b\right|\end{matrix}\right.\)

\(\Rightarrow cos\left(\overrightarrow{a},\overrightarrow{b}\right)=\frac{\overrightarrow{a}\overrightarrow{b}}{\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|}=\frac{\frac{b^2}{2}}{\left|a\right|.\left|b\right|}=\frac{\frac{b^2}{2}}{b^2}=\frac{1}{2}\)

\(\Rightarrow\left(\overrightarrow{a};\overrightarrow{b}\right)=60^0\)

Ta có: \(\overrightarrow {AB}  + \overrightarrow {BC}  = \overrightarrow {AC}  \Leftrightarrow \overrightarrow {BC}  = \overrightarrow b  - \overrightarrow a \)

Lại có: vecto \(\overrightarrow {BD} ,\overrightarrow {BC} \) cùng hướng và \(\left| {\overrightarrow {BD} } \right| = \frac{1}{3}\left| {\overrightarrow {BC} } \right|\)

\( \Rightarrow \overrightarrow {BD}  = \frac{1}{3}\overrightarrow {BC}  = \frac{1}{3}(\overrightarrow b  - \overrightarrow a )\)

Tương tự: vecto \(\overrightarrow {BE} ,\overrightarrow {BC} \) cùng hướng và \(\left| {\overrightarrow {BE} } \right| = \frac{2}{3}\left| {\overrightarrow {BC} } \right|\)

\( \Rightarrow \overrightarrow {BE}  = \frac{2}{3}\overrightarrow {BC}  = \frac{2}{3}(\overrightarrow b  - \overrightarrow a )\)

Ta có:

\(\overrightarrow {AB}  + \overrightarrow {BD}  = \overrightarrow {AD}  \Leftrightarrow \overrightarrow {AD}  = \overrightarrow a  + \frac{1}{3}(\overrightarrow b  - \overrightarrow a ) = \frac{2}{3}\overrightarrow a  + \frac{1}{3}\overrightarrow b \)

\(\overrightarrow {AB}  + \overrightarrow {BE}  = \overrightarrow {AE}  \Leftrightarrow \overrightarrow {AE}  = \overrightarrow a  + \frac{2}{3}(\overrightarrow b  - \overrightarrow a ) = \frac{1}{3}\overrightarrow a  + \frac{2}{3}\overrightarrow b \)

\(\overrightarrow{u}=2\overrightarrow{a}+3\overrightarrow{b}-5\overrightarrow{c}=\left(-30;21\right)\)