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Câu hỏi: Cho tam giác \(ABC\) đều có độ dài cạnh bằng \(3a\). Lấy điểm \(M\) thuộc cạnh \(BC\) sao cho \(MB = 2MC.\) Tích vô hướng của hai vectơ \(\overrightarrow {MA} \) và \(\overrightarrow {MC} \) bằng
A. \(\frac{{{a^2}}}{2}\)
B. \( - \frac{{{a^2}}}{2}\)
C. \({a^2}\)
D. \( - {a^2}\)
A. \(\frac{{{a^2}}}{2}\)
B. \( - \frac{{{a^2}}}{2}\)
C. \({a^2}\)
D. \( - {a^2}\)
Lời giải chi tiết
Ta có: \(\overrightarrow {MA} = \overrightarrow {MB} + \overrightarrow {BA} = - \overrightarrow {AB} - \frac{2}{3}\overrightarrow {BC} = - \frac{1}{3}\overrightarrow {AB} - \frac{2}{3}\overrightarrow {AC} \)
\(\overrightarrow {MC} = \frac{1}{3}\overrightarrow {CB} = \frac{1}{3}\left( {\overrightarrow {AB} - \overrightarrow {AC} } \right) = \frac{1}{3}\overrightarrow {AB} - \frac{1}{3}\overrightarrow {AC} \)
Ta có: \(\overrightarrow {MA} .\overrightarrow {MC} = \left( { - \frac{1}{3}\overrightarrow {AB} - \frac{2}{3}\overrightarrow {AC} } \right)\left( {\frac{1}{3}\overrightarrow {AB} - \frac{1}{3}\overrightarrow {AC} } \right)\)
\(\begin{array}{l} = - \frac{1}{9}{\overrightarrow {AB} ^2} + \frac{2}{9}{\overrightarrow {AC} ^2} - \frac{1}{9}\overrightarrow {AB} .\overrightarrow {AC} \\ = - \frac{1}{9}{\overrightarrow {AB} ^2} + \frac{2}{9}{\overrightarrow {AC} ^2} - \frac{1}{9}.AB.AC.\cos \widehat {BAC}\\ = - \frac{1}{9}.9{a^2} + \frac{2}{9}.9{a^2} - \frac{1}{9}.9{a^2}.\cos {60^ \circ }\\ = - {a^2} + 2{a^2} - {a^2}.\frac{1}{2} = \frac{{{a^2}}}{2}\end{array}\)
Ta có: \(\overrightarrow {MA} = \overrightarrow {MB} + \overrightarrow {BA} = - \overrightarrow {AB} - \frac{2}{3}\overrightarrow {BC} = - \frac{1}{3}\overrightarrow {AB} - \frac{2}{3}\overrightarrow {AC} \)
\(\overrightarrow {MC} = \frac{1}{3}\overrightarrow {CB} = \frac{1}{3}\left( {\overrightarrow {AB} - \overrightarrow {AC} } \right) = \frac{1}{3}\overrightarrow {AB} - \frac{1}{3}\overrightarrow {AC} \)
Ta có: \(\overrightarrow {MA} .\overrightarrow {MC} = \left( { - \frac{1}{3}\overrightarrow {AB} - \frac{2}{3}\overrightarrow {AC} } \right)\left( {\frac{1}{3}\overrightarrow {AB} - \frac{1}{3}\overrightarrow {AC} } \right)\)
\(\begin{array}{l} = - \frac{1}{9}{\overrightarrow {AB} ^2} + \frac{2}{9}{\overrightarrow {AC} ^2} - \frac{1}{9}\overrightarrow {AB} .\overrightarrow {AC} \\ = - \frac{1}{9}{\overrightarrow {AB} ^2} + \frac{2}{9}{\overrightarrow {AC} ^2} - \frac{1}{9}.AB.AC.\cos \widehat {BAC}\\ = - \frac{1}{9}.9{a^2} + \frac{2}{9}.9{a^2} - \frac{1}{9}.9{a^2}.\cos {60^ \circ }\\ = - {a^2} + 2{a^2} - {a^2}.\frac{1}{2} = \frac{{{a^2}}}{2}\end{array}\)
Đáp án A.