Cho ba vecto \(\overrightarrow{a}=\left(-2;1\right),\overrightarrow{b}=\left(1;3\right),\overrightarrow{c}=\left(-3;3\right)\). Hãy phân tích vecto \(\overrightarrow{c}\)thep 2 vecto \(\overrightarrow{a};\overrightarrow{b}\)
Cho các vecto \(\left|\overrightarrow{a}\right|=x,\left|\overrightarrow{b}\right|=y,\left|\overrightarrow{z}\right|=c\) và vecto a+b+3c=0. Tính \(A=\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{a}\)
\(\overrightarrow{a}+\overrightarrow{b}+3\overrightarrow{c}=\overrightarrow{0}\Leftrightarrow\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=-2\overrightarrow{c}\)
\(\Leftrightarrow\left(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\right)^2=\left(-2\overrightarrow{c}\right)^2\)
\(\Leftrightarrow\overrightarrow{a}^2+\overrightarrow{b}^2+\overrightarrow{c}^2+2\left(\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{a}\right)=4\overrightarrow{c}^2\)
\(\Leftrightarrow A=\dfrac{4x^2-\left(x^2+y^2+z^2\right)}{2}=\dfrac{3x^2-y^2-z^2}{2}\)
Cho hai vecto \(\overrightarrow{a};\overrightarrow{b}\) khác vecto 0. \(\left|\overrightarrow{a}\right|=4;\left|\overrightarrow{b}\right|=3;\left|\overrightarrow{a}-\overrightarrow{b}\right|=4\). Gọi \(\alpha\) là góc giữa hai vecto a vầ b. Chọn phát biểu đúng
A. \(\alpha\)= 60 độ B. \(\alpha\)= 30 độ C. \(\cos\alpha=\dfrac{1}{3}\) D\(\cos\alpha=\dfrac{3}{8}\)
\(\left|\overrightarrow{a}-\overrightarrow{b}\right|=4\)
⇒ \(\left(\overrightarrow{a}-\overrightarrow{b}\right)^2=16\)
⇒ 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
Cho 2 vecto \(\overrightarrow{a}\), \(\overrightarrow{b}\) vuông góc và \(\left|\overrightarrow{a}\right|=1\), \(\left|\overrightarrow{b}\right|=\sqrt{2}\). Chứng minh rằng 2 vecto sau vuông góc: \(\left(2\overrightarrow{a}-\overrightarrow{b}\right),\left(\overrightarrow{a}+\overrightarrow{b}\right)\).
\(\overrightarrow{a}\perp\overrightarrow{b}\Rightarrow\overrightarrow{a}.\overrightarrow{b}=0\)
\(\left(2\overrightarrow{a}-\overrightarrow{b}\right)\left(\overrightarrow{a}+\overrightarrow{b}\right)=2a^2+2\overrightarrow{a}.\overrightarrow{b}-\overrightarrow{a}.\overrightarrow{b}-b^2\)
\(=2a^2-b^2+\overrightarrow{a}.\overrightarrow{b}\)
\(=2.1-2+0=0\)
\(\Rightarrow\left(2\overrightarrow{a}-\overrightarrow{b}\right)\perp\left(\overrightarrow{a}+\overrightarrow{b}\right)\)
Cho hai vecto a và b sao cho \(\left|\overrightarrow{a}\right|=\sqrt{2},\left|\overrightarrow{b}\right|=2\) và hai vecto \(\overrightarrow{x}=\overrightarrow{a}+\overrightarrow{b},\overrightarrow{y}=\overrightarrow{2a}-\overrightarrow{b}\) vuông góc với nhau. Tính góc giữa hai vecto \(\overrightarrow{a},\overrightarrow{b}\)
\(\overrightarrow{x}\) ⊥ \(\overrightarrow{y}\)
⇒ \(\left(\overrightarrow{a}+\overrightarrow{b}\right)\left(\overrightarrow{2a}-\overrightarrow{b}\right)=0\). 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
⇒ \(\left(\overrightarrow{a};\overrightarrow{b}\right)=90^0\)
Cho hai vecto a;b khác vecto 0 thỏa mãn \(\overrightarrow{a}.\overrightarrow{b}=\dfrac{1}{2}\left|-\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\). Khi đó góc giữa hai vecto a và b là
Giả thiết => cos \(\left(\overrightarrow{a};\overrightarrow{b}\right)=\dfrac{1}{2}\)
⇒ \(\left(\overrightarrow{a};\overrightarrow{b}\right)=60^0\)
Tính \(\overrightarrow{a}.\overrightarrow{b}\) hả bạn?
\(\overrightarrow{a}.\overrightarrow{b}=\left|\overrightarrow{a}\right|.\left|\overrightarrow{b}\right|cos\left(\overrightarrow{a};\overrightarrow{b}\right)=2.\sqrt{3}.cos30^0=3\)
Đặt \(A=\left|\overrightarrow{a}+\overrightarrow{b}\right|\Rightarrow A^2=\left|\overrightarrow{a}\right|^2+\left|\overrightarrow{b}\right|^2+2\left|\overrightarrow{a}\right|.\left|\overrightarrow{b}\right|.cos\left(\overrightarrow{a};\overrightarrow{b}\right)\)
\(=2^2+3+2.2.\sqrt{3}.cos30^0=13\)
\(\Rightarrow\left|\overrightarrow{a}+\overrightarrow{b}\right|=\sqrt{13}\)
Cho 2 vecto không cùng phương \(\overrightarrow{a}\) và \(\overrightarrow{b}\)
CMR: \(\left|\overrightarrow{a}\right|-\left|\overrightarrow{b}\right|< \left|\overrightarrow{a}+\overrightarrow{b}\right|< \left|\overrightarrow{a}\right|+\left|\overrightarrow{b}\right|\)
Cho 2 vecto không cùng phương \(\overrightarrow{a}\) và \(\overrightarrow{b}\)
CMR : \(\left|\overrightarrow{a}\right|-\left|\overrightarrow{b}\right|< \left|\overrightarrow{a}+\overrightarrow{b}\right|< \left|\overrightarrow{a}\right|+\left|\overrightarrow{b}\right|\)
cho 2 vecto \(\overrightarrow{a},\overrightarrow{b}\) thoa man \(\left|\overrightarrow{a}\right|=4,\left|\overrightarrow{b}\right|=3\) và hai vecto \(\overrightarrow{u}=2\overrightarrow{a}+3\overrightarrow{b}\) và \(\overrightarrow{v}=-15\overrightarrow{a}+14\overrightarrow{b}\) vuông góc với nhau. Tính \(\left(\overrightarrow{a},\overrightarrow{b}\right)=???\)
\(u.v=0\Leftrightarrow\left(2a+3b\right)\left(-15a+14b\right)=0\)
\(\Leftrightarrow-30a^2+42b^2-17ab=0\)
\(\Leftrightarrow ab=\frac{-30.4^2+42.3^2}{17}=-6\)
\(\Rightarrow cos\left(a;b\right)=\frac{ab}{\left|a\right|\left|b\right|}=-\frac{6}{12}=-\frac{1}{2}\Rightarrow\left(a;b\right)=120^0\)