Đáp án:
Bài 3: 7
Bài 4: \(a)\,\,\sqrt 2 \), \(b)\,\,\left( {6; - 8} \right)\), c) 7;6, d) \( - \dfrac{1}{{\sqrt 2 }}\), đ) \(\overrightarrow d = 2\overrightarrow a + 5\overrightarrow b ,\,\,d = - 2\overrightarrow a + 4\overrightarrow b + \overrightarrow c ...\)
Giải thích các bước giải:
Bài 3:
ABCD là hình bình hành
\( \Rightarrow \overrightarrow {AB} = \overrightarrow {DC} \)
\(\begin{array}{l} \Rightarrow \left( { - 6;1} \right) = \left( {3 - a; - b} \right)\\ \Leftrightarrow \left\{ \begin{array}{l}3 - a = - 6\\ - b = 1\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}a = 9\\b = - 1\end{array} \right.\end{array}\)
Vậy \(a + 2b = 9 + 2\left( { - 1} \right) = 7\).
Bài 4:
a) \(\left| {\overrightarrow a } \right| = \sqrt {{1^2} + {{\left( { - 1} \right)}^2}} = \sqrt 2 \), \(\left| {\overrightarrow b } \right| = \sqrt {{0^2} + {1^2}} = 1\).
\( \Rightarrow \left| {\overrightarrow a } \right|.\left| {\overrightarrow b } \right| = \sqrt 2 .1 = \sqrt 2 \)
b) \(2\overrightarrow a - 3\overrightarrow b + \overrightarrow c \) \( = 2\left( {1; - 1} \right) - 3\left( {0;1} \right) + \left( {4; - 3} \right)\) \( = \left( {6; - 8} \right)\).
c) \(\overrightarrow a .\overrightarrow c = 1.4 - 1.\left( { - 3} \right) = 7\)
\(\overrightarrow a \left( {\overrightarrow b + \overrightarrow c } \right) = \left( {1; - 1} \right).\left( {4; - 2} \right) = 1.4 - 1.\left( { - 2} \right) = 6\)
d) \(\cos \left( {\overrightarrow a ;\overrightarrow b } \right) = \dfrac{{\overrightarrow a .\overrightarrow b }}{{\left| {\overrightarrow a } \right|.\left| {\overrightarrow b } \right|}} = \dfrac{{1.0 - 1.1}}{{\sqrt 2 }} = - \dfrac{1}{{\sqrt 2 }}\)
đ) Giả sử \(\overrightarrow d = x\overrightarrow a + y\overrightarrow b + z\overrightarrow c \)
\( \Rightarrow \left\{ \begin{array}{l} - 2 = x + 4z\\7 = - x + y - 3z\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}y + z = 5\\x = - 2 - 4z\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}y = 5 - z\\x = - 2 - 4z\end{array} \right.\)
\( \Rightarrow \overrightarrow d = \left( {2 - 4z} \right)\overrightarrow a + \left( {5 - z} \right)\overrightarrow b + z\overrightarrow c \)
Ví dụ chọn \(z = 0 \Rightarrow \overrightarrow d = 2\overrightarrow a + 5\overrightarrow b \), chọn \(z = 1 \Rightarrow d = - 2\overrightarrow a + 4\overrightarrow b + \overrightarrow c \),…
