Chương 2: TÍCH VÔ HƯỚNG CỦA HAI VECTƠ VÀ ỨNG DỤNG

\(3\overrightarrow{AP}-2\overrightarrow{AC}=\overrightarrow{0}\)

\(VT=3\left(\overrightarrow{AD}+\overrightarrow{DP}\right)-2\left(\overrightarrow{AD}+\overrightarrow{DC}\right)\)

\(=3\overrightarrow{AD}+3\overrightarrow{DP}-2\overrightarrow{AD}-2\overrightarrow{DC}\)

\(=\overrightarrow{AD}+3\overrightarrow{DP}-2\overrightarrow{DC}\)

\(=\overrightarrow{AD}+3\left(\overrightarrow{DC}+\overrightarrow{CP}\right)-2\overrightarrow{DC}\)

\(=\overrightarrow{AD}+3\overrightarrow{DC}+3\overrightarrow{CP}-2\overrightarrow{DC}\)

\(=\widehat{AD}+\overrightarrow{DC}+3.\dfrac{2}{3}\overrightarrow{CO}\)

\(=\overrightarrow{AD}+\overrightarrow{DC}+2.\dfrac{1}{2}\overrightarrow{CA}\)

\(=\overrightarrow{AD}+\overrightarrow{DC}+\overrightarrow{CA}\)

\(=\overrightarrow{AC}+\overrightarrow{CA}\)

\(=\overrightarrow{AA}=\overrightarrow{0}=VP\) (điều phải chứng minh)

a: \(\overrightarrow{AB}\cdot\overrightarrow{AD}=0\)

\(\overrightarrow{AB}\cdot\overrightarrow{BD}=\overrightarrow{AB}\left(\overrightarrow{AD}-\overrightarrow{AB}\right)=-AB^2=-a^2\)

b: \(=\overrightarrow{AB}\cdot\overrightarrow{BD}+\overrightarrow{AB}\cdot\overrightarrow{BC}+\overrightarrow{AD}\cdot\overrightarrow{BD}+\overrightarrow{AD}\cdot\overrightarrow{BC}\)

\(=-a^2-\overrightarrow{BA}\cdot\overrightarrow{BC}+\overrightarrow{DA}\cdot\overrightarrow{DB}+AD^2\)

\(=-0+DA\cdot DB\cdot cos45=a\cdot a\sqrt{2}\cdot\dfrac{\sqrt{2}}{2}=a^2\)

a: \(\overrightarrow{BA}\cdot\overrightarrow{BC}=BA\cdot BC\cdot cos60=\dfrac{1}{2}a^2\)

b: \(\overrightarrow{HB}\cdot\overrightarrow{BA}=\overrightarrow{HB}\left(\overrightarrow{HA}-\overrightarrow{HB}\right)=\overrightarrow{HB}\cdot\overrightarrow{HA}-HB^2=-HB^2=-\dfrac{1}{4}a^2\)

38.

Gọi I là trung điểm AB và G là trọng tâm tam giác ABC

\(\Rightarrow\left\{{}\begin{matrix}\overrightarrow{MA}+\overrightarrow{MB}=2\overrightarrow{MI}\\\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=\overrightarrow{0}\end{matrix}\right.\)

\(3\left|\overrightarrow{MA}+\overrightarrow{MB}\right|=2\left|\overrightarrow{MA}+\overrightarrow{MB}+\overrightarrow{MC}\right|\)

\(\Leftrightarrow3.\left|2\overrightarrow{MI}\right|=3\left|\overrightarrow{MG}+\overrightarrow{GA}+\overrightarrow{MG}+\overrightarrow{GB}+\overrightarrow{MG}+\overrightarrow{GC}\right|\)

\(\Leftrightarrow6\left|\overrightarrow{MI}\right|=2.\left|3\overrightarrow{MG}\right|\)

\(\Leftrightarrow6\left|\overrightarrow{MI}\right|=6\left|\overrightarrow{MG}\right|\)

\(\Leftrightarrow\left|\overrightarrow{MI}\right|=\left|\overrightarrow{MG}\right|\)

\(\Leftrightarrow MI=MG\)

\(\Rightarrow\) Tập hợp M là đường trung trực của đoạn thẳng IG

\(\overrightarrow{AB}.\overrightarrow{CD}=\overrightarrow{AB}\left(\overrightarrow{CB}+\overrightarrow{BA}+\overrightarrow{AD}\right)\)

\(=\overrightarrow{AB}.\overrightarrow{CB}+\overrightarrow{AB}.\overrightarrow{BA}+\overrightarrow{AB}.\overrightarrow{AD}\)

\(=0-\overrightarrow{AB}^2+0=-4a^2\)

\(AH=\dfrac{1}{2}AC\Rightarrow AC=2AH=4a\)

\(\Rightarrow AB=AC=BC=4a\)

\(\overrightarrow{AB}.\overrightarrow{BC}=AB.BC.cos\left(\overrightarrow{AB};\overrightarrow{BC}\right)=4a.4a.cos120^0=-8a^2\)

\(\overrightarrow{AH}.\overrightarrow{CA}=-\overrightarrow{AH}.\overrightarrow{AC}=-AH.AC.cos\left(\overrightarrow{AH};\overrightarrow{AC}\right)=-2a.4a.cos0^0=-8a^2\)

\(3\overrightarrow{MA}+2\overrightarrow{MB}=\overrightarrow{0}\) suy ra \(\dfrac{MB}{BA}=\dfrac{3}{5}\).

\(CPMQ\) là hình bình hành suy ra \(MP\) song song với \(CQ\), \(MQ\) song song với \(PC\). 

Xét tam giác \(ABC\) có \(MP\) song song với \(BC\) suy ra \(\dfrac{AP}{AC}=\dfrac{AM}{AB}=\dfrac{2}{5}\).

Suy ra \(\overrightarrow{AP}=\dfrac{2}{5}\overrightarrow{AC}\).

Tương tự ta cũng suy ra được \(\overrightarrow{BQ}=\dfrac{3}{5}\overrightarrow{BC}\). 

\(a\overrightarrow{NA}+b\overrightarrow{NQ}=\overrightarrow{0}\Leftrightarrow a\overrightarrow{AN}=b\overrightarrow{NQ}\) suy ra \(\overrightarrow{AQ}=\overrightarrow{AN}+\overrightarrow{NQ}=\overrightarrow{AN}+\dfrac{a}{b}\overrightarrow{AN}=\dfrac{a+b}{b}\overrightarrow{AN}\) 

do đó \(\overrightarrow{AN}=\dfrac{b}{a+b}\overrightarrow{AQ}\).

Để \(B,N,P\) thẳng hàng thì \(\overrightarrow{BN}\) và \(\overrightarrow{BP}\) cùng phương. Ta sẽ phân tích \(\overrightarrow{BN},\overrightarrow{BP}\) theo hai vectơ \(\overrightarrow{BA}\) và \(\overrightarrow{BC}\).

\(\overrightarrow{BN}=\overrightarrow{BA}+\overrightarrow{AN}=\overrightarrow{BA}+\dfrac{b}{a+b}\overrightarrow{AQ}=\overrightarrow{BA}+\dfrac{b}{a+b}\left(\overrightarrow{AB}+\overrightarrow{BQ}\right)\)

\(=\overrightarrow{BA}-\dfrac{b}{a+b}\overrightarrow{BA}+\dfrac{3b}{5\left(a+b\right)}\overrightarrow{BC}=\dfrac{a}{a+b}\overrightarrow{BA}+\dfrac{3b}{5\left(a+b\right)}\overrightarrow{BC}\). 

\(\overrightarrow{BP}=\overrightarrow{BA}+\overrightarrow{AP}=\overrightarrow{BA}+\dfrac{2}{5}\overrightarrow{AC}=\overrightarrow{BA}+\dfrac{2}{5}\left(\overrightarrow{AB}+\overrightarrow{BC}\right)=\dfrac{3}{5}\overrightarrow{BA}+\dfrac{2}{5}\overrightarrow{BC}\).

\(\overrightarrow{BN},\overrightarrow{BP}\) cùng phương nên 

\(\dfrac{\dfrac{a}{a+b}}{\dfrac{3}{5}}=\dfrac{\dfrac{3b}{5\left(a+b\right)}}{\dfrac{2}{5}}\Leftrightarrow\dfrac{2a}{5\left(a+b\right)}=\dfrac{9b}{25\left(a+b\right)}\Leftrightarrow10a=9b\).

Vì \(\left(a,b\right)=1\), \(a,b\) nguyên nên \(a=9,b=10\).

Vậy \(a+b=19\).

Câu 2:

AB=BC=CD=DA=4a

\(AC=BD=\sqrt{2\cdot\left(4a\right)^2}=4a\sqrt{2}\)

\(\overrightarrow{AB}\cdot\overrightarrow{AC}=AB\cdot AC\cdot cos\widehat{BAC}\)

\(=4a\cdot4a\sqrt{2}\cdot\dfrac{\sqrt{2}}{2}=16a^2\)

Vì AB vuông góc AD

nên vecto AB*vecto AD=0

vecto AB*vecto BC

=-vecto BC*vecto BA

=0

vecto AC*vecto CB

=-vecto CA*vecto CB

=-CA*CB*cos góc ACB

\(=-4a\sqrt{2}\cdot4a\cdot\dfrac{\sqrt{2}}{2}=-16a^2\)

vecto AD*vecto DC

=-vecto DA*vecto DC

=0