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Câu 4:
Áp dụng định lý Pytago
\(BC^2=AB^2+AC^2\Rightarrow BC=2\)
Ta có:
\(\overrightarrow{CA}.\overrightarrow{BC}=-\overrightarrow{CA}.\overrightarrow{CB}=-\dfrac{CA^2+CB^2-AB^2}{2}=-\dfrac{2+4-2}{2}=-2\)
Câu 5:
Gọi M là trung điểm BC
\(\overrightarrow{AM}=\dfrac{1}{2}\left(\overrightarrow{AB}+\overrightarrow{AC}\right)\)
Mà: \(\overrightarrow{AG}=\dfrac{2}{3}\overrightarrow{AM}=\dfrac{1}{3}\left(\overrightarrow{AB}+\overrightarrow{AC}\right)\)
Câu 6:
\(\left|\overrightarrow{a}-\overrightarrow{b}\right|=3\)
\(a^2+b^2-2\overrightarrow{a}.\overrightarrow{b}=9\)
\(\overrightarrow{a}.\overrightarrow{b}=\dfrac{1^2+2^2-9}{2}=-2\)
Câu 7:
\(\left|\overrightarrow{AB}-\overrightarrow{AD}+\overrightarrow{CD}\right|=\left|\overrightarrow{DB}+\overrightarrow{CD}\right|\)
\(=\left|\overrightarrow{DB}-\overrightarrow{DC}\right|=\left|\overrightarrow{CB}\right|=BC=a\)
\(\overrightarrow{a}\) ⊥\(\overrightarrow{b}\)
=>-1*2+2*m=0
=>2m-2=0
=>2m=2
=>m=1
=>Chọn D
\(\overrightarrow{a}\cdot\overrightarrow{b}=\left|\overrightarrow{a}\right|\cdot\left|\overrightarrow{b}\right|\cdot cos\left(\overrightarrow{a};\overrightarrow{b}\right)\)
\(=3\cdot2\cdot cos60=6\cdot\frac12=3\)
=>Chọn C
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)
Gọi G là trọng tâm tam giác \(\Rightarrow\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=0\)
\(\overrightarrow{MA}^2+\overrightarrow{MA}.\overrightarrow{MB}+\overrightarrow{MA}.\overrightarrow{MC}=0\)
\(\Leftrightarrow\overrightarrow{MA}\left(\overrightarrow{MA}+\overrightarrow{MB}+\overrightarrow{MC}\right)=0\)
\(\Leftrightarrow\overrightarrow{MA}\left(\overrightarrow{MG}+\overrightarrow{GA}+\overrightarrow{MG}+\overrightarrow{GB}+\overrightarrow{MG}+\overrightarrow{GC}\right)=0\)
\(\Leftrightarrow3\overrightarrow{MA}.\overrightarrow{MG}=0\)
\(\Rightarrow\) M thuộc đường tròn đường kính AG
Bán kính: \(R=\dfrac{1}{2}AG=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{\sqrt{3}}{2}=\dfrac{\sqrt{3}}{6}\)
\(\overrightarrow{a}.\overrightarrow{b}=\left|\overrightarrow{a}\right|.\left|\overrightarrow{b}\right|.cos\left(\overrightarrow{a};\overrightarrow{b}\right)\)
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/ \(\overrightarrow{a}.\overrightarrow{b}=5.6.cos120^0=-15\)
\(\left|4\cdot\overrightarrow{a}-3\cdot\overrightarrow{b}\right|=\sqrt{13}\)
=>\((4\vec{a} - 3\vec{b})^2 = 13\)
=>\(16\vert{}\vec{a}\vert{}^2-24(\vec{a}\cdot\vec{b})+9\vert{}\vec{b}\vert{}^2=13\)
=>\(16\cdot1^2-24(\vec{a}\cdot\vec{b})+9\cdot^2=13\)
=>\(25 - 24(\vec{a} \cdot \vec{b}) = 13\)
=>\(24\cdot\overrightarrow{a}\cdot\overrightarrow{b}=12\)
=>\(\overrightarrow{a}\cdot\overrightarrow{b}=\frac12\)
\(\overrightarrow{c}\) ⊥\(\left(\overrightarrow{a}-\overrightarrow{b}\right)\)
=>\(\vec{c} \cdot (\vec{a} - \vec{b}) = 0\)
=>\((x\vec{a}+y\vec{b})\cdot(\vec{a}-\vec{b})=0\)
=>\(x\vert{}\vec{a}\vert{}^2-x(\vec{a}\cdot\vec{b})+y(\vec{a}\cdot\vec{b})-y\vert{}\vec{b}\vert{}^2=0\)
=>\(x - \frac{1}{2}x + \frac{1}{2}y - y = 0\)
=>x=y
\(\vec{c} = x\vec{a} + x\vec{b} = x(\vec{a} + \vec{b})\)
\(\vert{}\vec{c}\vert{}^2 = x^2(\vec{a} + \vec{b})^2 = 1\)
=>\(x^2(\vert{}\vec{a}\vert{}^2+2\vec{a}\cdot\vec{b}+\vert{}\vec{b}\vert{}^2)=1\)
=>\(x^2 \left(1 + 2 \cdot \frac{1}{2} + 1\right) = 1\)
=>\(x^2=\frac13\)
=>\(x=\pm\frac{\sqrt3}{3}\)
Khi \(x=\frac{\sqrt3}{3}\) thì \(y=x=\frac{\sqrt3}{3}\)
Khi \(x=-\frac{\sqrt3}{3}\) thì \(y=x=-\frac{\sqrt3}{3}\)