Cho A(4;3), B(2;7), C(-3;-8) a. Tính vectơ AB× vectơ AC b. Tìm tạo độ trực tâm H của ∆ABC 07/08/2021 Bởi Savannah Cho A(4;3), B(2;7), C(-3;-8) a. Tính vectơ AB× vectơ AC b. Tìm tạo độ trực tâm H của ∆ABC
Đáp án: $\begin{array}{l}a)\,\,\overrightarrow {AB} .\overrightarrow {AC} = – 30.\\b)\,\,H\left( {13;\,\,0} \right).\end{array}$ Giải thích các bước giải: Cho \(A\left( {4;\,\,3} \right),\,\,\,B\left( {2;\,\,7} \right),\,\,C\left( { – 3; – 8} \right)\) a) Tính \(\overrightarrow {AB} .\overrightarrow {AC} .\) Ta có: \(\left\{ \begin{array}{l}\overrightarrow {AB} = \left( { – 2;\,\,4} \right)\\\overrightarrow {AC} = \left( { – 7;\,\, – 11} \right)\\\overrightarrow {BC} = \left( { – 5;\,\, – 15} \right)\end{array} \right..\) \( \Rightarrow \overrightarrow {AB} .\overrightarrow {AC} = \left( { – 2} \right).\left( { – 7} \right) + 4.\left( { – 11} \right) = – 30.\) b) Tìm tọa độ trực tâm \(H\) của\(\Delta ABC.\) Gọi \(H\left( {a;\,\,b} \right)\) là trực tâm \(\Delta ABC.\) Khi đó ta có: \(\begin{array}{l}\left\{ \begin{array}{l}\overrightarrow {AH} \bot \overrightarrow {BC} \\\overrightarrow {BH} \bot \overrightarrow {AC} \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\overrightarrow {AH} .\overrightarrow {BC} = 0\\\overrightarrow {BH} .\overrightarrow {AC} = 0\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}\left( {a – 4;\,\,b – 3} \right).\left( { – 5; – 15} \right) = 0\\\left( {a – 2;\,\,b – 7} \right)\left( { – 7; – 11} \right) = 0\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}5\left( {a – 4} \right) + 15\left( {b – 3} \right) = 0\\7\left( {a – 2} \right) + 11\left( {b – 7} \right) = 0\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}a – 4 + 3b – 9 = 0\\7a – 14 + 11b – 77 = 0\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}a + 3b = 13\\7a + 11b = 91\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = 13\\b = 0\end{array} \right. \Rightarrow H\left( {13;\,\,0} \right).\end{array}\) Bình luận
Đáp án:
$\begin{array}{l}
a)\,\,\overrightarrow {AB} .\overrightarrow {AC} = – 30.\\
b)\,\,H\left( {13;\,\,0} \right).
\end{array}$
Giải thích các bước giải:
Cho \(A\left( {4;\,\,3} \right),\,\,\,B\left( {2;\,\,7} \right),\,\,C\left( { – 3; – 8} \right)\)
Ta có: \(\left\{ \begin{array}{l}\overrightarrow {AB} = \left( { – 2;\,\,4} \right)\\\overrightarrow {AC} = \left( { – 7;\,\, – 11} \right)\\\overrightarrow {BC} = \left( { – 5;\,\, – 15} \right)\end{array} \right..\)
\( \Rightarrow \overrightarrow {AB} .\overrightarrow {AC} = \left( { – 2} \right).\left( { – 7} \right) + 4.\left( { – 11} \right) = – 30.\)
Gọi \(H\left( {a;\,\,b} \right)\) là trực tâm \(\Delta ABC.\)
Khi đó ta có:
\(\begin{array}{l}\left\{ \begin{array}{l}\overrightarrow {AH} \bot \overrightarrow {BC} \\\overrightarrow {BH} \bot \overrightarrow {AC} \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\overrightarrow {AH} .\overrightarrow {BC} = 0\\\overrightarrow {BH} .\overrightarrow {AC} = 0\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}\left( {a – 4;\,\,b – 3} \right).\left( { – 5; – 15} \right) = 0\\\left( {a – 2;\,\,b – 7} \right)\left( { – 7; – 11} \right) = 0\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}5\left( {a – 4} \right) + 15\left( {b – 3} \right) = 0\\7\left( {a – 2} \right) + 11\left( {b – 7} \right) = 0\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}a – 4 + 3b – 9 = 0\\7a – 14 + 11b – 77 = 0\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}a + 3b = 13\\7a + 11b = 91\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = 13\\b = 0\end{array} \right. \Rightarrow H\left( {13;\,\,0} \right).\end{array}\)