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\(\overrightarrow{x}\) ⊥ \(\overrightarrow{y}\)
⇒ \(\left(\overrightarrow{a}+\overrightarrow{b}\right)\left(\overrightarrow{2a}-\overrightarrow{b}\right)=0\). 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
⇒ \(\left(\overrightarrow{a};\overrightarrow{b}\right)=90^0\)
\(\overrightarrow{a}\perp\overrightarrow{b}\Rightarrow\overrightarrow{a}.\overrightarrow{b}=0\)
\(\left(2\overrightarrow{a}-\overrightarrow{b}\right)\left(\overrightarrow{a}+\overrightarrow{b}\right)=2a^2+2\overrightarrow{a}.\overrightarrow{b}-\overrightarrow{a}.\overrightarrow{b}-b^2\)
\(=2a^2-b^2+\overrightarrow{a}.\overrightarrow{b}\)
\(=2.1-2+0=0\)
\(\Rightarrow\left(2\overrightarrow{a}-\overrightarrow{b}\right)\perp\left(\overrightarrow{a}+\overrightarrow{b}\right)\)
\(\overrightarrow{a}+\overrightarrow{b}+3\overrightarrow{c}=\overrightarrow{0}\Leftrightarrow\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=-2\overrightarrow{c}\)
\(\Leftrightarrow\left(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\right)^2=\left(-2\overrightarrow{c}\right)^2\)
\(\Leftrightarrow\overrightarrow{a}^2+\overrightarrow{b}^2+\overrightarrow{c}^2+2\left(\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{a}\right)=4\overrightarrow{c}^2\)
\(\Leftrightarrow A=\dfrac{4x^2-\left(x^2+y^2+z^2\right)}{2}=\dfrac{3x^2-y^2-z^2}{2}\)
\(\overrightarrow{m}=2\left(1;2\right)+3\left(3;4\right)=\left(2;4\right)+\left(9;12\right)=\left(11;16\right)\)
\(\left|\overrightarrow{a}-\overrightarrow{b}\right|=4\)
⇒ \(\left(\overrightarrow{a}-\overrightarrow{b}\right)^2=16\)
⇒ 16 + 9 - 2\(\overrightarrow{a}.\overrightarrow{b}\) = 16
⇒ \(2\overrightarrow{a}.\overrightarrow{b}=9\)
⇒ cosα = \(\dfrac{9}{2.4.3}\)
⇒ cos α = \(\dfrac{3}{8}\)
Vậy chọn D
Giả thiết => cos \(\left(\overrightarrow{a};\overrightarrow{b}\right)=\dfrac{1}{2}\)
⇒ \(\left(\overrightarrow{a};\overrightarrow{b}\right)=60^0\)
\(\left(a+2b\right)^2=28\Leftrightarrow a^2+4b^2+4ab=28\)
\(\Rightarrow ab=\frac{28-4^2-4.3^2}{4}=-6\)
\(\Rightarrow cos\left(a;b\right)=-\frac{6}{4.3}=-\frac{1}{2}\Rightarrow\left(a;b\right)=120^0\)
(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
\(\left|\overrightarrow{a}-2\cdot\overrightarrow{b}\right|=\sqrt{15}\)
=>\(\left(\overrightarrow{a}-2\cdot\overrightarrow{b}\right)\left(\overrightarrow{a}-2\cdot\overrightarrow{b}\right)=15\)
=>\(\overrightarrow{a}\cdot\overrightarrow{a}-4\cdot\overrightarrow{a}\cdot\overrightarrow{b}+4\cdot\overrightarrow{b}\cdot\overrightarrow{b}=15\)
=>\(\left(\left|\overrightarrow{a}\right|\right)^2-4\cdot\overrightarrow{a}\cdot\overrightarrow{b}+4\cdot\left(\overrightarrow{b}\right)^2=15\)
=>\(1^2+4\cdot2^2-4\cdot\overrightarrow{a}\cdot\overrightarrow{b}=15\)
=>\(4\cdot\overrightarrow{a}\cdot\overrightarrow{b}=1+16-15=2\)
=>\(\overrightarrow{a}\cdot\overrightarrow{b}=\frac12\)
b: \(\left(\overrightarrow{a}+\overrightarrow{b}\right)\left(2k\cdot\overrightarrow{a}-\overrightarrow{b}\right)\)
\(=2k\cdot\overrightarrow{a}\cdot\overrightarrow{a}-\overrightarrow{a}\cdot\overrightarrow{b}+2k\cdot\overrightarrow{a}\cdot\overrightarrow{b}-\overrightarrow{b}\cdot\overrightarrow{b}\)
\(=2k\cdot\left(\left|\overrightarrow{a}\right|\right)^2+\overrightarrow{a}\cdot\overrightarrow{b}\left(2k-1\right)-\left(\overrightarrow{b}\right)^2\)
\(=2k\cdot1^2+\left(2k-1\right)\cdot\frac12-2^2=2k+k-\frac12-4=3k-\frac92\)
\(\left(\overrightarrow{a}+\overrightarrow{b}\right)\left(\overrightarrow{a}+\overrightarrow{b}\right)=\left(\left|\overrightarrow{a}\right|\right)^2+2\cdot\overrightarrow{a}\cdot\overrightarrow{b}+\left(\left|\overrightarrow{b}\right|\right)^2\)
\(=1^2+2^2+2\cdot\frac12=5+1=6\)
=>\(\left|\overrightarrow{a}+\overrightarrow{b}\right|=\sqrt6\)
\(\left(2k\cdot\overrightarrow{a}-\overrightarrow{b}\right)^2=4k^2\cdot\left(\left|\overrightarrow{a}\right|\right)^2-2\cdot2k\cdot\overrightarrow{a}\cdot\overrightarrow{b}+\left(\overrightarrow{b}\right)^2\)
\(=4k^2\cdot1-4k\cdot\frac12+4=4k^2-2k+4\)
=>\(\left|2k\cdot\overrightarrow{a}-\overrightarrow{b}\right|=\sqrt{4k^2-2k+4}\)
\(cos\left(\left(\overrightarrow{a}+\overrightarrow{b}\right);\left(2k\cdot\overrightarrow{a}-\overrightarrow{b}\right)\right)=cos60^0=\frac12\)
=>\(\frac{3k-4,5}{\sqrt{6\left(4k^2-2k+4\right)}}=\frac12\)
=>\(\sqrt{\frac{\left(3k-4,5\right)^2}{6\left(4k^2-2k+4\right)}}=\frac12\)
=>\(\frac{\left(3k-4,5\right)^2}{6\left(4k^2-2k+4\right)}=\frac14\)
=>\(6\left(4k^2-2k+4\right)=4\left(3k-4,5\right)^2\)
=>\(4\left(9k^2-27k+20,25\right)=6\left(4k^2-2k+4\right)\)
=>\(36k^2-108k+81=24k^2-12k+24\)
=>\(12k^2-96k+57=0\)
=>\(4k^2-32k+19=0\)
=>\(k=\frac{8\pm3\sqrt5}{2}\)
Tính \(\overrightarrow{a}.\overrightarrow{b}\) hả bạn?
\(\overrightarrow{a}.\overrightarrow{b}=\left|\overrightarrow{a}\right|.\left|\overrightarrow{b}\right|cos\left(\overrightarrow{a};\overrightarrow{b}\right)=2.\sqrt{3}.cos30^0=3\)
Tính \(\left|\overrightarrow{a}+\overrightarrow{b}\right|\)
Đặt \(A=\left|\overrightarrow{a}+\overrightarrow{b}\right|\Rightarrow A^2=\left|\overrightarrow{a}\right|^2+\left|\overrightarrow{b}\right|^2+2\left|\overrightarrow{a}\right|.\left|\overrightarrow{b}\right|.cos\left(\overrightarrow{a};\overrightarrow{b}\right)\)
\(=2^2+3+2.2.\sqrt{3}.cos30^0=13\)
\(\Rightarrow\left|\overrightarrow{a}+\overrightarrow{b}\right|=\sqrt{13}\)