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a) Theo giả thiết \(\overrightarrow{a}=\overrightarrow{b}\ne\overrightarrow{0}\) nên giả sử \(\overrightarrow{a}=m\overrightarrow{b}\) suy ra:
\(\overrightarrow{a}=m\overrightarrow{a}\Leftrightarrow\left(1-m\right)\overrightarrow{a}=\overrightarrow{0}\).
\(\Leftrightarrow1-m=0\) (vì \(\overrightarrow{a}\ne\overrightarrow{0}\) ).
\(\Leftrightarrow m=1\).
b) Nếu \(\overrightarrow{a}=-\overrightarrow{b};\overrightarrow{a}\ne\overrightarrow{0}\).
Giả sử \(\overrightarrow{a}=m\overrightarrow{b}\Leftrightarrow\overrightarrow{a}=-m\overrightarrow{a}\)\(\Leftrightarrow\overrightarrow{a}\left(1+m\right)=\overrightarrow{0}\)
\(\Leftrightarrow1+m=0\)\(\Leftrightarrow m=-1\).
c) Do \(\overrightarrow{a}\) , \(\overrightarrow{b}\) cùng hướng nên: \(m>0\).
Mặt khác: \(\overrightarrow{a}=m\overrightarrow{b}\Leftrightarrow\left|\overrightarrow{a}\right|=\left|m\right|.\left|\overrightarrow{b}\right|\)
\(\Leftrightarrow20=5.\left|m\right|\)\(\Leftrightarrow\left|m\right|=4\)
\(\Leftrightarrow m=\pm4\).
Do m > 0 nên m = 4.
d) Do \(\overrightarrow{a},\overrightarrow{b}\) ngược hướng nên m < 0.
\(\left|\overrightarrow{a}\right|=\left|m\right|.\left|\overrightarrow{b}\right|\)\(\Leftrightarrow15=\left|m\right|.3\)\(\Leftrightarrow\left|m\right|=5\)\(\Leftrightarrow m=\pm5\).
Do m < 0 nên m = -5.
e) \(\overrightarrow{a}=\overrightarrow{0};\overrightarrow{b}\ne\overrightarrow{0}\) nên\(\overrightarrow{0}=m.\overrightarrow{b}\). Suy ra m = 0.
g) \(\overrightarrow{a}\ne\overrightarrow{0};\overrightarrow{b}=\overrightarrow{0}\) nên \(\overrightarrow{a}=m.\overrightarrow{0}=\overrightarrow{0}\). Suy ra không tồn tại giá trị m thỏa mãn.
h) \(\overrightarrow{a}=\overrightarrow{0};\overrightarrow{b}=\overrightarrow{0}\) nên \(\overrightarrow{0}=m.\overrightarrow{0}\). Suy ra mọi \(m\in R\) đều thỏa mãn.
\(\left|\overrightarrow{a}+\overrightarrow{b}\right|^2=\left(\overrightarrow{a}+\overrightarrow{b}\right)\left(\overrightarrow{a}+\overrightarrow{b}\right)\)
\(=\left|\overrightarrow{a}\right|^2+\left|\overrightarrow{b}\right|^2+2\overrightarrow{a}.\overrightarrow{b}\)
\(=5^2+12^2+2.5.12.cos\left(\overrightarrow{a},\overrightarrow{b}\right)\)
\(=169+120cos\left(\overrightarrow{a},\overrightarrow{b}\right)=13^2\)
Suy ra: \(cos\left(\overrightarrow{a};\overrightarrow{b}\right)=0\).
\(\overrightarrow{a}\left(\overrightarrow{a}+\overrightarrow{b}\right)=\left(\overrightarrow{a}\right)^2+\overrightarrow{a}.\overrightarrow{b}=5^2+5.12.0=25\).
Mặt khác \(\overrightarrow{a}\left(\overrightarrow{a}+\overrightarrow{b}\right)=\left|\overrightarrow{a}\right|.\left|\overrightarrow{a}+\overrightarrow{b}\right|.cos\left(\overrightarrow{a},\overrightarrow{a}+\overrightarrow{b}\right)\)
\(=5.13.cos\left(\overrightarrow{a},\overrightarrow{a}+\overrightarrow{b}\right)\).
Vì vậy \(25=5.13.cos\left(\overrightarrow{a},\overrightarrow{a}+\overrightarrow{b}\right)\).
\(cos\left(\overrightarrow{a},\overrightarrow{a}+\overrightarrow{b}\right)=\dfrac{5}{13}\).
Vậy góc giữa hai véc tơ \(\overrightarrow{a}\) và \(\overrightarrow{a}+\overrightarrow{b}\) là \(\alpha\) sao cho \(cos\alpha=\dfrac{5}{13}\).
\(BC=AD=\sqrt{AC^2-AB^2}=2a\)
a/ \(T=\left|3\overrightarrow{AB}-4\overrightarrow{BC}\right|\Rightarrow T^2=9AB^2+16BC^2-24\overrightarrow{AB}.\overrightarrow{BC}\)
\(=9a^2+64a^2=73a^2\Rightarrow T=a\sqrt{73}\)
b/ \(T^2=4AB^2+9BC^2+12.\overrightarrow{BA}.\overrightarrow{BC}=4AB^2+9BC^2=40a^2\)
\(\Rightarrow T=2a\sqrt{10}\)
c/ \(T=\left|\overrightarrow{AD}+3\overrightarrow{BC}\right|=\left|\overrightarrow{AD}+3\overrightarrow{AD}\right|=\left|4\overrightarrow{AD}\right|=4AD=8a\)
d/ \(T=\left|2\overrightarrow{DC}-3\overrightarrow{DC}\right|=\left|-\overrightarrow{DC}\right|=CD=AB=a\)
a) Biểu thị các vectơ \(\overset{\rightarrow}{A B} , \&\text{nbsp}; \overset{\rightarrow}{B C} , \&\text{nbsp}; \overset{\rightarrow}{A C}\) theo \(\overset{⃗}{a} , \&\text{nbsp}; \overset{⃗}{b}\)
Ta có:
Suy ra:
1. Vectơ \(\overset{\rightarrow}{A B}\)
\(\overset{\rightarrow}{A B} = \overset{\rightarrow}{A E} + \overset{\rightarrow}{E B} = \overset{⃗}{a} + \overset{⃗}{b}\)
2. Vectơ \(\overset{\rightarrow}{B C}\)
Trên đoạn \(B C\), ta có chia đều:
\(B E = E F = F C \Rightarrow C \&\text{nbsp};\text{c} \overset{ˊ}{\text{a}} \text{ch}\&\text{nbsp}; B \&\text{nbsp};\text{m}ộ\text{t}\&\text{nbsp};đ\text{o}ạ\text{n}\&\text{nbsp}; 3 \&\text{nbsp};\text{ph} \overset{ˋ}{\hat{\text{a}}} \text{n}\&\text{nbsp};\text{b} \overset{ˋ}{\overset{ }{\text{a}}} \text{ng}\&\text{nbsp};\text{nhau}\)
Từ B đến C:
\(\overset{\rightarrow}{B C} = \overset{\rightarrow}{B E} + \overset{\rightarrow}{E F} + \overset{\rightarrow}{F C}\)
Ta biết:
\(\overset{\rightarrow}{B E} = - \overset{\rightarrow}{E B} = - \overset{⃗}{b}\)
Đồng thời:
\(\overset{\rightarrow}{E F} = \overset{\rightarrow}{F C} = \overset{\rightarrow}{B E} = - \overset{⃗}{b}\)
Vậy:
\(\overset{\rightarrow}{B C} = - \overset{⃗}{b} - \overset{⃗}{b} - \overset{⃗}{b} = - 3 \overset{⃗}{b}\)
3. Vectơ \(\overset{\rightarrow}{A C}\)
\(\overset{\rightarrow}{A C} = \overset{\rightarrow}{A E} + \overset{\rightarrow}{E C} = \overset{⃗}{a} + \left(\right. \overset{\rightarrow}{E B} + \overset{\rightarrow}{B C} \left.\right)\)
Nhưng \(\overset{\rightarrow}{E B} = \overset{⃗}{b}\), \(\overset{\rightarrow}{B C} = - 3 \overset{⃗}{b}\), nên:
\(\overset{\rightarrow}{E C} = \overset{⃗}{b} - 3 \overset{⃗}{b} = - 2 \overset{⃗}{b}\)
Do đó:
\(\overset{\rightarrow}{A C} = \overset{⃗}{a} - 2 \overset{⃗}{b}\)
Kết quả phần a)
\(\boxed{\overset{\rightarrow}{A B} = \overset{⃗}{a} + \overset{⃗}{b} , \overset{\rightarrow}{B C} = - 3 \overset{⃗}{b} , \overset{\rightarrow}{A C} = \overset{⃗}{a} - 2 \overset{⃗}{b}}\)
b) Tính \(\overset{\rightarrow}{A B} \cdot \overset{\rightarrow}{A C}\)
Ta có:
\(\overset{\rightarrow}{A B} = \overset{⃗}{a} + \overset{⃗}{b} , \overset{\rightarrow}{A C} = \overset{⃗}{a} - 2 \overset{⃗}{b}\)
Tích vô hướng:
\(\left(\right. \overset{⃗}{a} + \overset{⃗}{b} \left.\right) \cdot \left(\right. \overset{⃗}{a} - 2 \overset{⃗}{b} \left.\right) = \overset{⃗}{a} \cdot \overset{⃗}{a} - 2 \overset{⃗}{a} \cdot \overset{⃗}{b} + \overset{⃗}{b} \cdot \overset{⃗}{a} - 2 \overset{⃗}{b} \cdot \overset{⃗}{b}\)
Gom nhóm:
\(= \mid \overset{⃗}{a} \mid^{2} - \overset{⃗}{a} \cdot \overset{⃗}{b} - 2 \mid \overset{⃗}{b} \mid^{2}\)
Ta biết:
\(\overset{⃗}{a} \cdot \overset{⃗}{b} = \mid \overset{⃗}{a} \mid \textrm{ } \mid \overset{⃗}{b} \mid cos 120^{\circ} = 5 \cdot 2 \cdot \left(\right. - \frac{1}{2} \left.\right) = - 5\)
Thay vào:
\(\overset{\rightarrow}{A B} \cdot \overset{\rightarrow}{A C} = 25 - \left(\right. - 5 \left.\right) - 2 \cdot 4\) \(= 25 + 5 - 8 = 22\)
Kết quả phần b)
\(\boxed{\overset{\rightarrow}{A B} \cdot \overset{\rightarrow}{A C} = 22}\)
TÍCH CHO MÌNH ĐC KO