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a: \(\left|\overrightarrow{AB}+\overrightarrow{AD}+\overrightarrow{AC}\right|=2\cdot AC=2\cdot5=10\)
b: \(\left|\overrightarrow{AM}+\overrightarrow{AN}\right|=\left|\dfrac{\overrightarrow{AB}+\overrightarrow{AC}}{2}+\dfrac{\overrightarrow{AD}+\overrightarrow{AC}}{2}\right|\)
\(=\left|\dfrac{3\cdot\overrightarrow{AC}}{2}\right|=\dfrac{3}{2}AC=\dfrac{3}{2}\cdot5=\dfrac{15}{2}=7.5\)
** M là trung điểm của AB đúng không bạn?
a.
\(|\overrightarrow{AM}+\overrightarrow{AB}|=|\frac{1}{2}\overrightarrow{AB}+\overrightarrow{AB}|=\frac{3}{2}|\overrightarrow{AB}|=\frac{3}{2}.3a=\frac{9a}{2}\)
b.
\(|\overrightarrow{AB}+\overrightarrow{CD}|=|\overrightarrow{AB}+\overrightarrow{BA}|=|\overrightarrow{0}|=0\)
c.Trên $CD$ lấy $K$ sao cho $CK=a$. Khi đó:
\(|\overrightarrow{DN}+\overrightarrow{BN}|=|\overrightarrow{DN}+\overrightarrow{KD}|=|\overrightarrow{KN}|=KN=\sqrt{a^2+a^2}=\sqrt{2}a\)
\(\left|\overrightarrow{AB}+\overrightarrow{AD}\right|=AC=5a\)
a: vecto AB-vecto AD
=vecto DA+vecto AB
=vecto DB
-vecto CD-veco BC
=vecto CB-vecto CD
=vecto DC+vecto CB=vecto DB
=>vecto AB+vecto CD=vecto AD-vecto BC
b: \(\overrightarrow{AB}-\overrightarrow{AC}=\overrightarrow{CA}+\overrightarrow{AB}=\overrightarrow{CB}\)
\(\overrightarrow{CD}-\overrightarrow{BD}=\overrightarrow{CD}+\overrightarrow{DB}=\overrightarrow{CB}\)
Do đó: \(\overrightarrow{AB}-\overrightarrow{AC}=\overrightarrow{CD}-\overrightarrow{BD}\)
=>\(\overrightarrow{AB}-\overrightarrow{CD}=\overrightarrow{AC}-\overrightarrow{BD}\)
c: \(\overrightarrow{AB}-\overrightarrow{AD}=\overrightarrow{DA}+\overrightarrow{AB}=\overrightarrow{DB}\)
\(\overrightarrow{CB}-\overrightarrow{CD}=\overrightarrow{DC}+\overrightarrow{CB}=\overrightarrow{DB}\)
Do đó: \(\overrightarrow{AB}-\overrightarrow{AD}=\overrightarrow{CB}-\overrightarrow{CD}\)
=>\(\overrightarrow{AB}+\overrightarrow{CD}=\overrightarrow{AD}+\overrightarrow{CB}\)
Xét ΔADB có
\(cosA=\dfrac{AB^2+AD^2-DB^2}{2\cdot AB\cdot AD}\)
=>\(\dfrac{a^2+9a^2-DB^2}{2\cdot a\cdot3a}=\dfrac{1}{2}\)
=>\(10a^2-DB^2=3a^2\)
=>\(DB=a\sqrt{7}\)
Xét ΔABD có
\(cosABD=\dfrac{BA^2+BD^2-AD^2}{2\cdot BA\cdot BD}\)
\(=\dfrac{9a^2+7a^2-a^2}{2\cdot3a\cdot a\sqrt{7}}=\dfrac{15a^2}{6a^2\cdot\sqrt{7}}=\dfrac{15}{6\sqrt{7}}=\dfrac{5}{2\sqrt{7}}\)
=>\(cosCDB=\dfrac{5}{2\sqrt{7}}\)(do \(\widehat{ABD}=\widehat{CDB}\) vì AB//CD)
Xét ΔCDB có \(cosCDB=\dfrac{DB^2+DC^2-BC^2}{2\cdot DB\cdot DC}\)
=>\(\dfrac{5}{2\sqrt{7}}=\dfrac{7a^2+a^2-BC^2}{2\cdot a\sqrt{7}\cdot a}\)
=>\(\dfrac{8a^2-BC^2}{2a^2\sqrt{7}}=\dfrac{5}{2\sqrt{7}}\)
=>\(\dfrac{8a^2-BC^2}{a^2}=5\)
=>\(8a^2-BC^2=5a^2\)
=>\(BC^2=3a^2\)
=>\(BC=a\sqrt{3}\)
ta có
\(M=\left|\overrightarrow{AD}+2\overrightarrow{BC}\right|=\left|\overrightarrow{AD}+2\overrightarrow{AD}\right|=3\left|\overrightarrow{AD}\right|=3AD=3a\)
Vậy độ dài của \(\left|\overrightarrow{AD}+2\overrightarrow{BC}\right|=3a\)