[tex]\overrightarrow{AC}=4\overrightarrow{AM}=>\overrightarrow{AB}=\frac{\overrightarrow{BC}-4\overrightarrow{BM}}{-3}<=>3\overrightarrow{BA}+\overrightarrow{BC}=4\overrightarrow{MB}<=>16MB^2=9BA^2+BC^2+6\overrightarrow{BA}\overrightarrow{BC}<=>16MB^2=9.16+16<=>MB^2=10=>MB=\sqrt{10}[/tex]
[tex]\frac{BN}{BD}=\frac{\frac{2}{3BD}}{BD}=\frac{2}{3BD^2}=\frac{1}{48}=>BN=\frac{1}{48}BD[/tex]
[tex]\overrightarrow{MN}=\overrightarrow{BN}-\overrightarrow{BM}=\frac{1}{48}\overrightarrow{BD}-\overrightarrow{BM}<=> MN^2=\frac{1}{2304}BD^2+BM^2-\frac{1}{24}\overrightarrow{BD}\overrightarrow{BM}=\frac{721}{72}-\frac{1}{24}cos(\overrightarrow{BD},\overrightarrow{BM}).\left | \overrightarrow{BD} \right |\left | \overrightarrow{BM} \right |=\frac{721}{72}-\frac{\sqrt{5}}{3}cos(\overrightarrow{BD},\overrightarrow{BM})[/tex]
[tex]\overrightarrow{AC}=4\overrightarrow{AM}<=>\overrightarrow{AD}=\frac{\overrightarrow{DC}-4\overrightarrow{DM}}{-3}<=>3\overrightarrow{DA}-\overrightarrow{DC}=4\overrightarrow{MD}<=>16MD^2=9DA^2+DC^2-6\overrightarrow{DA}\overrightarrow{DC}<=>MD=\sqrt{10}[/tex]
[tex]cos(\overrightarrow{BD},\overrightarrow{BM})=\frac{BD^2+BM^2-DM^2}{2BD.BM}=\frac{32+10-10}{2.\sqrt{2}.4.\sqrt{10}}=\frac{2}{\sqrt{5}}[/tex]
=> [tex]MN^2=\frac{721}{72}-\frac{\sqrt{5}}{3}cos(\overrightarrow{BD},\overrightarrow{BM})=\frac{721}{72}-\frac{\sqrt{5}}{3}.\frac{2}{\sqrt{5}}=\frac{673}{72}=>MN=\sqrt{\frac{673}{72}}[/tex]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
