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a, \(AC=\dfrac{AB}{sin45^o}=\dfrac{a}{\dfrac{\sqrt{2}}{2}}=a\sqrt{2}\)
\(\overrightarrow{AB}.\overrightarrow{AC}=AB.AC.cos\widehat{BAC}=a.a\sqrt{2}.cos45^o=a^2\)
b, \(\left(\overrightarrow{AB}+\overrightarrow{AD}\right)\left(\overrightarrow{BD}+\overrightarrow{BC}\right)=\overrightarrow{AC}\left(\overrightarrow{BD}+\overrightarrow{BC}\right)\)
\(=\overrightarrow{AC}.\overrightarrow{BD}+\overrightarrow{AC}.\overrightarrow{BC}\)
\(=AC.BD.cos90^o+AC.AD.cos45^o\)
\(=a\sqrt{2}.a\sqrt{2}.0+a\sqrt{2}.a.\dfrac{\sqrt{2}}{2}=a^2\)
c, \(\overrightarrow{AB}.\overrightarrow{BD}=AB.BD.cos135^o=-a.a\sqrt{2}.\dfrac{\sqrt{2}}{2}=-a^2\)
d, \(\left(\overrightarrow{AC}-\overrightarrow{AB}\right)\left(2\overrightarrow{AD}-\overrightarrow{AB}\right)=\overrightarrow{BC}.\left(\overrightarrow{AD}+\overrightarrow{BD}\right)\)
\(=\overrightarrow{BC}.\overrightarrow{AD}+\overrightarrow{BC}.\overrightarrow{BD}\)
\(=AD^2+BC.BD.cos45^o\)
\(=a^2+a.a\sqrt{2}.\dfrac{\sqrt{2}}{2}=2a^2\)
e, \(\left(\overrightarrow{AB}+\overrightarrow{AC}+\overrightarrow{AD}\right)\left(\overrightarrow{DA}+\overrightarrow{DB}+\overrightarrow{DC}\right)\)
\(=\left(\overrightarrow{AC}+\overrightarrow{AC}\right)\left(\overrightarrow{DB}+\overrightarrow{DB}\right)\)
\(=4.\overrightarrow{AC}.\overrightarrow{DB}=4.AC.DB.cos90^o=0\)
a) \(\left| {\overrightarrow a + \overrightarrow b } \right| = \left| {\overrightarrow a } \right| + \left| {\overrightarrow b } \right| \Leftrightarrow {\left| {\overrightarrow a + \overrightarrow b } \right|^2} = {\left( {\left| {\overrightarrow a } \right| + \left| {\overrightarrow b } \right|} \right)^2}\)
\( \Leftrightarrow {\left( {\overrightarrow a + \overrightarrow b } \right)^2} = {\left( {\left| {\overrightarrow a } \right| + \left| {\overrightarrow b } \right|} \right)^2} \Leftrightarrow {\left( {\overrightarrow a } \right)^2} + 2\overrightarrow a .\overrightarrow b + {\left( {\overrightarrow b } \right)^2} = {\left| {\overrightarrow a } \right|^2} + 2.\left| {\overrightarrow a } \right|.\left| {\overrightarrow b } \right| + {\left| {\overrightarrow b } \right|^2}\)
\( \Leftrightarrow {\left| {\overrightarrow a } \right|^2} + 2\overrightarrow a .\overrightarrow b + {\left| {\overrightarrow b } \right|^2} = {\left| {\overrightarrow a } \right|^2} + 2.\left| {\overrightarrow a } \right|.\left| {\overrightarrow b } \right| + {\left| {\overrightarrow b } \right|^2}\)
\( \Leftrightarrow 2\overrightarrow a .\overrightarrow b = 2\left| {\overrightarrow a } \right|.\left| {\overrightarrow b } \right|\)
\( \Leftrightarrow 2\left| {\overrightarrow a } \right|.\left| {\overrightarrow b } \right|\cos \left( {\overrightarrow a ,\overrightarrow b } \right) = 2\left| {\overrightarrow a } \right|.\left| {\overrightarrow b } \right|\)
\( \Leftrightarrow \cos \left( {\overrightarrow a ,\overrightarrow b } \right) = 1 \Leftrightarrow \left( {\overrightarrow a ,\overrightarrow b } \right) = 0^\circ \)
Vậy \(\left| {\overrightarrow a + \overrightarrow b } \right| = \left| {\overrightarrow a } \right| + \left| {\overrightarrow b } \right| \Leftrightarrow \overrightarrow a , \,\overrightarrow b \) cùng hướng.
b) \(\left| {\overrightarrow a + \overrightarrow b } \right| = \left| {\overrightarrow a - \overrightarrow b } \right| \Leftrightarrow {\left| {\overrightarrow a + \overrightarrow b } \right|^2} = {\left| {\overrightarrow a - \overrightarrow b } \right|^2}\)
\( \Leftrightarrow {\left( {\overrightarrow a + \overrightarrow b } \right)^2} = {\left( {\overrightarrow a - \overrightarrow b } \right)^2}\)
\( \Leftrightarrow {\left( {\overrightarrow a } \right)^2} + 2\overrightarrow a .\overrightarrow b + {\left( {\overrightarrow b } \right)^2} = {\left( {\overrightarrow a } \right)^2} - 2\overrightarrow a .\overrightarrow b + {\left( {\overrightarrow b } \right)^2}\)
\( \Leftrightarrow 2\overrightarrow a .\overrightarrow b = - 2\overrightarrow a .\overrightarrow b \Leftrightarrow 4\overrightarrow a .\overrightarrow b = 0\)
\( \Leftrightarrow \overrightarrow a .\overrightarrow b = 0 \Leftrightarrow \left( {\overrightarrow a ,\overrightarrow b } \right) = 90^\circ \)
Vậy \(\left| {\overrightarrow a + \overrightarrow b } \right| = \left| {\overrightarrow a - \overrightarrow b } \right| \Leftrightarrow \overrightarrow a ,\overrightarrow b \) vuông góc với nhau.
Do tam giác ABC vuông tại A và \(\widehat{B}=30^o\) \(\Rightarrow C=60^o\)
\(\Rightarrow\left(\overrightarrow{AB},\overrightarrow{BC}\right)=150^o;\)\(\left(\overrightarrow{BA},\overrightarrow{BC}\right)=30^o;\left(\overrightarrow{AC},\overrightarrow{CB}\right)=120^o\)
\(\left(\overrightarrow{AB},\overrightarrow{AC}\right)=90^o;\left(\overrightarrow{BC},\overrightarrow{BA}\right)=30^o\).Do vậy:
a) \(\cos\left(\overrightarrow{AB},\overrightarrow{BC}\right)+\sin\left(\overrightarrow{BA},\overrightarrow{BC}\right)+\tan\frac{\left(\overrightarrow{AC},\overrightarrow{CB}\right)}{2}\)
\(=\cos150^o+\sin30^o+\tan60^o\)
\(=-\frac{\sqrt{3}}{2}+\frac{1}{2}+\sqrt{3}\)
\(=\frac{\sqrt{3}+1}{2}\)
b) \(\sin\left(\overrightarrow{AB},\overrightarrow{AC}\right)+\cos\left(\overrightarrow{BC},\overrightarrow{AB}\right)+\cos\left(\overrightarrow{CA},\overrightarrow{BA}\right)\)
\(=\sin90^o+\cos30^o+\cos0^o\)
\(=1+\frac{\sqrt{3}}{2}\)
\(=\frac{2+\sqrt{3}}{2}\)
Lời giải:
Xét hai vecto bất kỳ \(\overrightarrow{AB}, \overrightarrow{CD}\). Kẻ vecto $\overrightarrow{CT}$ sao cho $\overrightarrow{CT}=\overrightarrow{BA}$
Ta có:
\(|\overrightarrow{AB}+\overrightarrow{CD}|=|\overrightarrow{TC}+\overrightarrow{CD}|=|\overrightarrow{TD}|\)
\(|\overrightarrow{AB}|+|\overrightarrow{CD}|=|\overrightarrow{TC}|+|\overrightarrow{CD}|\)
Mà theo bđt tam giác thì:
\(|\overrightarrow{TC}+\overrightarrow{CD}|\geq |\overrightarrow{TD}|\Rightarrow |\overrightarrow{AB}|+\overrightarrow{CD}|\geq |\overrightarrow{AB}+\overrightarrow{CD}|\)
Dấu "=" xảy ra khi \(T, C,D\) thẳng hàng và $C$ nằm giữa $T,D$
$\Leftrightarrow \overrightarrow{TC}, \overrightarrow{CD}$ cùng hướng
$\Leftrightarrow \overrightarrow{AB}, \overrightarrow{CD}$ cùng hướng
Vậy với $\overrightarrow{a}, \overrightarrow{b}$ bất kỳ thì $|\overrightarrow{a}|+|\overrightarrow{b}|\geq |\overrightarrow{a}+\overrightarrow{b}|$. Dấu "=" xảy ra khi $\overrightarrow{a}, \overrightarrow{b}$ cùng hướng.
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Áp dụng vào bài toán:
\(|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}|\leq |\overrightarrow{a}+\overrightarrow{b}|+|\overrightarrow{c}|\leq |\overrightarrow{a}|+|\overrightarrow{b}|+|\overrightarrow{c}|\)
Dấu "=" xảy ra khi \(\overrightarrow{a}, \overrightarrow{b}\) cùng hướng và \(\overrightarrow{a}+\overrightarrow{b}, \overrightarrow{c}\) cùng hướng
\(\Leftrightarrow \overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}\) cùng hướng
a)
\(\overrightarrow{u}=\overrightarrow{AB}+\overrightarrow{DC}+\overrightarrow{BD}+\overrightarrow{CA}\)
\(=\overrightarrow{AB}+\overrightarrow{BD}+\overrightarrow{DC}+\overrightarrow{CA}\)
\(=\overrightarrow{AD}+\overrightarrow{DA}=\overrightarrow{0}\).
b)
\(\overrightarrow{v}=\overrightarrow{AB}+\overrightarrow{CD}+\overrightarrow{BC}+\overrightarrow{DA}\)
\(=\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CD}+\overrightarrow{DA}\)
\(=\overrightarrow{AC}+\overrightarrow{CA}=\overrightarrow{0}\).
a) \(\overrightarrow{u}=3\overrightarrow{a}+2\overrightarrow{b}-4\overrightarrow{c}=3\left(2;1\right)+2\left(3;-4\right)-4\left(-7;2\right)\)
\(=\left(6;3\right)+\left(6;-8\right)-\left(-28;8\right)\)
\(=\left(6+6+28;3-8-8\right)=\left(40;-13\right)\).
b) \(\overrightarrow{x}+\overrightarrow{a}=\overrightarrow{b}-\overrightarrow{c}\Leftrightarrow\overrightarrow{x}=\overrightarrow{b}-\overrightarrow{c}-\overrightarrow{a}\)
\(\Leftrightarrow\overrightarrow{x}=\left(3;-4\right)-\left(-7;2\right)-\left(2;1\right)\)
\(\Leftrightarrow\overrightarrow{x}=\left(3+7-2;-4-2-1\right)\)
\(\Leftrightarrow\overrightarrow{x}=\left(8;-7\right)\).
c) Có \(\overrightarrow{c}\left(-7;2\right)=k\overrightarrow{a}+h\overrightarrow{b}=k\left(2;1\right)+h\left(3;-4\right)\)
\(=\left(2k+3h;k-4h\right)\).
Từ đó suy ra: \(\left\{{}\begin{matrix}2k+3h=-7\\k-4h=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}k=-2\\h=-1\end{matrix}\right.\).
\(\)vectơ \(\overrightarrow c = \frac{{\left| {\overrightarrow a } \right|}}{{\left| {\overrightarrow b } \right|}}.\overrightarrow b \) có độ dài gấp \(\frac{{\left| {\overrightarrow a } \right|}}{{\left| {\overrightarrow b } \right|}}\) lần vectơ \(\overrightarrow b \) và cùng hướng với vectơ \(\overrightarrow b \)
+) Nếu hai vectơ \(\overrightarrow a \) và \(\overrightarrow b \) cùng hướng thì hai vectơ \(\overrightarrow a \) và \(\overrightarrow c \)cùng hướng và ngược lại
+) \(\left| {\overrightarrow c } \right| = \left| {\frac{{\left| {\overrightarrow a } \right|}}{{\left| {\overrightarrow b } \right|}}.\overrightarrow b } \right| = \frac{{\left| {\overrightarrow a } \right|}}{{\left| {\overrightarrow b } \right|}}.\left| {\overrightarrow b } \right| = \left| {\overrightarrow a } \right|\). Suy ra hai vectơ \(\overrightarrow a \) và \(\overrightarrow c \)có cùng độ dài
a) Áp dụng quy tắc ba điểm ta có:
\(\overrightarrow a + \overrightarrow b = \overrightarrow {AB} + \overrightarrow {BC} = \overrightarrow {AC} \);
\(\overrightarrow b + \overrightarrow a = \overrightarrow {AE} + \overrightarrow {EC} = \overrightarrow {AC} \)
\( \Rightarrow \overrightarrow a + \overrightarrow b = \overrightarrow b + \overrightarrow a \)
b) Áp dụng quy tắc ba điểm ta có:
\(\left( {\overrightarrow a + \overrightarrow b } \right) + \overrightarrow c = \left( {\overrightarrow {AB} + \overrightarrow {BC} } \right) + \overrightarrow {CD} = \overrightarrow {AC} + \overrightarrow {CD} = \overrightarrow {AD} \)
\(\overrightarrow a + \left( {\overrightarrow b + \overrightarrow c } \right) = \overrightarrow {AB} + \left( {\overrightarrow {BC} + \overrightarrow {CD} } \right) = \overrightarrow {AB} + \overrightarrow {BD} = \overrightarrow {AD} \)
\( \Rightarrow \left( {\overrightarrow a + \overrightarrow b } \right) + \overrightarrow c = \overrightarrow a + \left( {\overrightarrow b + \overrightarrow c } \right)\)