Giải bài 61 trang 105 SBT toán 10 - Cánh diều

Đề bài

Cho tam giác ABC đều cạnh a. Các điểm M, N lần lượt thuộc các tia BC và CA thoả mãn \(BM = \frac{1}{3}BC,CN = \frac{5}{4}CA\). Tính:

a) \(\overrightarrow {AB} .\overrightarrow {AC} ,\overrightarrow {AM} .\overrightarrow {BN} \)

b) MN

Lời giải chi tiết

a) Ta có:

* \(\overrightarrow {AB} .\overrightarrow {AC}  = AB.AC.\cos \widehat {BAC} = a.a.\cos {60^0} = \frac{{{a^2}}}{2}\)

* \(\overrightarrow {AM} .\overrightarrow {BN}  = \left( {\overrightarrow {CM}  - \overrightarrow {CA} } \right)\left( {\overrightarrow {CN}  - \overrightarrow {CB} } \right) = \overrightarrow {CM} .\overrightarrow {CN}  - \overrightarrow {CM} .\overrightarrow {CB}  - \overrightarrow {CA} .\overrightarrow {CN}  + \overrightarrow {CA} .\overrightarrow {CB} \)

Ta có: + \(\overrightarrow {CM} .\overrightarrow {CN}  = CM.CN.\cos \widehat {MCN} = \frac{{2a}}{3}.\frac{{5a}}{4}.\cos {60^0} = \frac{{5{a^2}}}{{12}}\)

           + \(\overrightarrow {CM} .\overrightarrow {CB}  = \frac{2}{3}\overrightarrow {CB} .\overrightarrow {CB}  = \frac{2}{3}B{C^2} = \frac{{2{a^2}}}{3}\)

           + \(\overrightarrow {CA} .\overrightarrow {CN}  = \overrightarrow {CA} .\frac{5}{4}\overrightarrow {CA}  = \frac{5}{4}A{C^2} = \frac{{5{a^2}}}{4}\)

           + \(\overrightarrow {CA} .\overrightarrow {CB}  = CA.CB.\cos \widehat {ACB} = a.a.\cos {60^0} = \frac{{{a^2}}}{2}\)

\( \Rightarrow \overrightarrow {AM} .\overrightarrow {BN}  = \overrightarrow {CM} .\overrightarrow {CN}  - \overrightarrow {CM} .\overrightarrow {CB}  - \overrightarrow {CA} .\overrightarrow {CN}  + \overrightarrow {CA} .\overrightarrow {CB}  = \frac{{5{a^2}}}{{12}} - \frac{{2{a^2}}}{3} - \frac{{5{a^2}}}{4} + \frac{{{a^2}}}{2} =  - {a^2}\)

Vậy \(\overrightarrow {AB} .\overrightarrow {AC}  = \frac{{{a^2}}}{2}\), \(\overrightarrow {AM} .\overrightarrow {BN}  =  - {a^2}\)

b) Ta có: \(M{N^2} = {\left( {\overrightarrow {MN} } \right)^2} = {\left( {\overrightarrow {MB}  + \overrightarrow {BC}  + \overrightarrow {CN} } \right)^2} = {\left( { - \frac{1}{3}\overrightarrow {BC}  + \overrightarrow {BC}  + \frac{5}{4}\overrightarrow {CA} } \right)^2}\) 

                      \( = {\left( {\frac{2}{3}\overrightarrow {BC}  + \frac{5}{4}\overrightarrow {CA} } \right)^2} = \frac{4}{9}B{C^2} + \frac{{25}}{{16}}A{C^2} + \frac{5}{3}\overrightarrow {BC} .\overrightarrow {CA} \)

                      \( = \frac{{289}}{{144}}{a^2} - \frac{5}{3}\overrightarrow {CB} .\overrightarrow {CA}  = \frac{{289}}{{144}}{a^2} - \frac{5}{3}.CB.CA.\cos \widehat {BCA}\) \( = \frac{{289}}{{144}}{a^2} - \frac{5}{6}{a^2} = \frac{{169{a^2}}}{{144}}\)

\( \Rightarrow M{N^2} = \frac{{169{a^2}}}{{144}} \Rightarrow MN = \frac{{13a}}{{12}}\)