by Cho $ \left| \overrightarrow{a} \right|+\left| \overrightarrow{b} \right|=2$ và $ \left( \overrightarrow{a};\overrightarrow{b} \right)={{0}^{0}},\left| 2\overrightarrow{a}+\overrightarrow{b} \right|=3$ . Tìm $ \left| \overrightarrow{a} \right|$ và $ \left| \overrightarrow{b} \right|$ .
A. $ \left| \overrightarrow{a} \right|=\frac{5}{3};\left| \overrightarrow{b} \right|=\frac{1}{3}. $
B. $ \left| \overrightarrow{a} \right|=\frac{4}{5};\left| \overrightarrow{b} \right|=\frac{6}{5}. $
C. $ \left| \overrightarrow{a} \right|=\left| \overrightarrow{b} \right|=1. $
D. $ \left| \overrightarrow{a} \right|=\frac{1}{2};\left| \overrightarrow{b} \right|=\frac{3}{2}. $
Hướng dẫn
$ \left| 2\overrightarrow{a}+\overrightarrow{b} \right|=3\Leftrightarrow 4{{\overrightarrow{a}}^{2}}+{{\overrightarrow{b}}^{2}}+4\overrightarrow{a}. \overrightarrow{b}=9$
$ \Leftrightarrow 4{{\left| {\vec{a}} \right|}^{2}}+4\left| {\vec{a}} \right|\left| {\vec{b}} \right|\cos \left( \vec{a};\vec{b} \right)+{{\left| {\vec{b}} \right|}^{2}}=9$
$ \Leftrightarrow 4{{\left( 2-\left| {\vec{b}} \right| \right)}^{2}}+4\left( 2-\left| {\vec{b}} \right| \right)\left| {\vec{b}} \right|+{{\left| {\vec{b}} \right|}^{2}}=9$
$ \Leftrightarrow {{\left| {\vec{b}} \right|}^{2}}-8\left| {\vec{b}} \right|+7=0\Leftrightarrow \left[ \begin{array}{l}
\left| {\vec{b}} \right|=1\Rightarrow \left| {\vec{a}} \right|=1 \\
\left| {\vec{b}} \right|=7\Rightarrow \left| {\vec{a}} \right|=-5\,\,(loai)
\end{array} \right. $
Vậy, $ \left| {\vec{a}} \right|=1,\left| {\vec{b}} \right|=1. $
Chọn đáp án C.
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