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\(3BM=7CM=7\left(BC-BM\right)\Rightarrow10BM=7BC\)
\(\Rightarrow BM=\dfrac{7}{10}BC\Rightarrow\overrightarrow{BM}=\dfrac{7}{10}\overrightarrow{BC}\)
Ta có:
\(\overrightarrow{AM}=\overrightarrow{AB}+\overrightarrow{BM}=\overrightarrow{AB}+\dfrac{7}{10}\overrightarrow{BC}=\overrightarrow{AB}+\dfrac{7}{10}\left(\overrightarrow{BA}+\overrightarrow{AC}\right)=\overrightarrow{AB}-\dfrac{7}{10}\overrightarrow{AB}+\dfrac{7}{10}\overrightarrow{AC}\)
\(\Rightarrow\overrightarrow{AM}=\dfrac{3}{10}\overrightarrow{AB}+\dfrac{7}{10}\overrightarrow{AC}\)
Lời giải:
Theo đề thì $\overrightarrow{3BM}=7\overrightarrow{MC}=-7\overrightarrow{CM}$
Lại có:
$\overrightarrow{AM}=\overrightarrow{AB}+\overrightarrow{BM}$
$\Rightarrow 3\overrightarrow{AM}=3\overrightarrow{AB}+3\overrightarrow{BM}=3\overrightarrow{AB}-7\overrightarrow{CM}(1)$
$\overrightarrow{AM}=\overrightarrow{AC}+\overrightarrow{CM}$
$\Rightarrow 7\overrightarrow{AM}=7\overrightarrow{AC}+7\overrightarrow{CM}(2)$
Từ $(1);(2)\Rightarrow 10\overrightarrow{AM}=3\overrightarrow{AB}+7\overrightarrow{AC}$
$\Rightarrow \overrightarrow{AM}=\frac{3}{10}\overrightarrow{AB}+\frac{7}{10}\overrightarrow{AC}$
Dùng kết quả: Nếu B, C, M thẳng hàng và A M → = x A B → + y A C → thì x + y = 1 để loại các phương án A, B, D.
Đáp án C
\(5\overrightarrow{JB}=2\overrightarrow{JC}=2\left(\overrightarrow{JB}+\overrightarrow{BC}\right)=2\overrightarrow{JB}+2\overrightarrow{BC}\)
\(\Rightarrow\overrightarrow{JB}=\dfrac{2}{3}\overrightarrow{BC}=2\overrightarrow{BA}+2\overrightarrow{AC}\Rightarrow\overrightarrow{BJ}=2\overrightarrow{AB}-2\overrightarrow{AC}\)
\(\Rightarrow\overrightarrow{AJ}=\overrightarrow{AB}+\overrightarrow{BJ}=\overrightarrow{AB}+2\overrightarrow{AB}-2\overrightarrow{AC}=3\overrightarrow{AB}-2\overrightarrow{AC}\)
Ta có M B → = 1 3 M C → ⇔ 3 M B → = M C → ⇔ 3 B M → = C M →
A M → = A B → + B M → ⇒ 3 A M → = 3 A B → + 3 B M → ( 1 ) A M → = A C → + C M → ( 2 )
Lấy (1) trừ (2) ta được :
2 A M → = 3 A B → + 3 B M → − A C → + C M → = 3 A B → − A C → + ( 3 B M → − C M → ) = 3 A B → − A C → + 0 → = 3 A B → − A C → ⇒ A M → = 3 2 A B → − 1 2 A C → = 3 2 u → − 1 2 v →
Đáp án A
a) Ta có:
\(\overrightarrow{AM}=\overrightarrow{AB}+\overrightarrow{BM}\)
\(=\overrightarrow{AB}+k\overrightarrow{BC}\)
\(=\overrightarrow{AB}+k\left(\overrightarrow{AC}-\overrightarrow{AB}\right)\)
\(=\left(1-k\right)\overrightarrow{AB}+k\overrightarrow{AC}\)
b) \(\overrightarrow{NP}=\overrightarrow{AP}-\overrightarrow{AN}\)
\(=\dfrac{2}{3}\overrightarrow{AC}-\dfrac{3}{4}\overrightarrow{AB}\)
Để \(AM\perp NP\)
\(\Rightarrow\overrightarrow{AM}.\overrightarrow{NP}=\overrightarrow{0}\)
\(\Rightarrow\left[\left(1-k\right)\overrightarrow{AB}+k\overrightarrow{AC}\right]\left(-\dfrac{3}{4}\overrightarrow{AB}+\dfrac{2}{3}\overrightarrow{AC}\right)=\overrightarrow{0}\)
\(\Leftrightarrow\dfrac{3\left(k-1\right)}{4}AB^2+\dfrac{2k}{3}AC^2+\dfrac{2\left(1-k\right)}{3}\overrightarrow{AB}.\overrightarrow{AC}-\dfrac{3k}{4}\overrightarrow{AB}.\overrightarrow{AC}=\overrightarrow{0}\)
\(\Leftrightarrow\dfrac{3\left(k-1\right)}{4}AB^2+\dfrac{2k}{3}AB^2+\dfrac{1-k}{3}AB^2-\dfrac{3k}{8}AB^2=0\)
\(\Leftrightarrow AB^2\left[\dfrac{3\left(k-1\right)}{4}+\dfrac{2k}{3}+\dfrac{1-k}{3}-\dfrac{3k}{8}\right]=0\)
\(\Leftrightarrow18\left(k-1\right)+16k+8\left(1-k\right)-9k=0\left(AB>0\right)\)
\(\Leftrightarrow17k=10\)
\(\Leftrightarrow k=\dfrac{10}{17}\)