جواب تمرین صفحه 84 درس 3 هندسه دوازدهم (بردارها)

الف 

\(\begin{array}{l}\left. \begin{array}{l}\overrightarrow a = (2\;,\; - 1\;,\;2)\;,\;\overrightarrow b = (1\;,\;0\;,\;0)\\\\ \Rightarrow \overrightarrow a \cdot \overrightarrow b = 2 \times 1 + ( - 1) \times 0 + 2 \times 0 = 2\\\\ \Rightarrow {b^2} = {1^2} + {0^2} + {0^2} = 1\end{array} \right\}\\\\ \Rightarrow \overrightarrow {a'} = \frac{{\overrightarrow a \cdot \overrightarrow b }}{{{b^2}}}\overrightarrow b = \frac{2}{1}(1\;,\;0\;,\;0) = (2\;,\;0\;,\;0)\end{array}\)

ب 

\(\begin{array}{l}\left. \begin{array}{l}\overrightarrow a = (2\;,\;3\;,\;1)\;,\;\overrightarrow b = (3\;,\;2\;,\;1)\\\\ \Rightarrow \overrightarrow a \cdot \overrightarrow b = 2 \times 3 + 3 \times 2 + 1 \times 1 = 13\\\\ \Rightarrow {b^2} = {3^2} + {2^2} + {1^2} = 14\end{array} \right\}\\\\ \Rightarrow \overrightarrow {a'} = \frac{{\overrightarrow a \cdot \overrightarrow b }}{{{b^2}}}\overrightarrow b = \frac{{13}}{{14}}(3\;,\;2\;,\;1)\\\\ = (\frac{{39}}{{14}}\;,\;\frac{{26}}{{14}}\;,\;\frac{{13}}{{14}})\end{array}\)

پ 

\(\begin{array}{l}\left. \begin{array}{l}\overrightarrow a = (1\;,\;1\;,\;0)\;,\;\overrightarrow b = ( - 1\;,\;2\;,\;4)\\\\ \Rightarrow \overrightarrow a \cdot \overrightarrow b = 1 \times ( - 1) + 1 \times 2 + 0 \times 4 = 1\\\\ \Rightarrow {b^2} = {( - 1)^2} + {2^2} + {4^2} = 21\end{array} \right\}\\\\ \Rightarrow \overrightarrow {a'} = \frac{{\overrightarrow a \cdot \overrightarrow b }}{{{b^2}}}\overrightarrow b = \frac{1}{{21}}( - 1\;,\;2\;,\;4)\\\\ = ( - \frac{1}{{21}}\;,\;\frac{2}{{21}}\;,\;\frac{4}{{21}})\end{array}\)

\(\begin{array}{l}\left| {\overrightarrow a } \right| = 2\;,\;\left| {\overrightarrow b } \right| = 1\;,\;\left| {\overrightarrow c } \right| = 3\\\\\overrightarrow a + \overrightarrow b + \overrightarrow c = \overrightarrow O \;\mathop \Rightarrow \limits^{ \cdot \overrightarrow a } \\\\\overrightarrow a \cdot \overrightarrow a + \overrightarrow b \cdot \overrightarrow a + \overrightarrow c \cdot \overrightarrow a = {a^2} + \overrightarrow a \cdot \overrightarrow b + \overrightarrow a \cdot \overrightarrow c = 0\;\;\;(1)\\\\\overrightarrow a + \overrightarrow b + \overrightarrow c = \overrightarrow O \;\mathop \Rightarrow \limits^{ \cdot \overrightarrow b } \\\\\overrightarrow a \cdot \overrightarrow b + \overrightarrow b \cdot \overrightarrow b + \overrightarrow c \cdot \overrightarrow b = \overrightarrow a \cdot \overrightarrow b + {b^2} + \overrightarrow b \cdot \overrightarrow c = 0\;\;\;(2)\\\\\overrightarrow a + \overrightarrow b + \overrightarrow c = \overrightarrow O \;\mathop \Rightarrow \limits^{ \cdot \overrightarrow c } \;\\\\\overrightarrow a \cdot \overrightarrow c + \overrightarrow b \cdot \overrightarrow c + \overrightarrow c \cdot \overrightarrow c = \overrightarrow a \cdot \overrightarrow b + \overrightarrow a \cdot \overrightarrow b + {c^2} = 0\;\;\;(3)\\\\\mathop \Rightarrow \limits^{(1)\;,\;(2)\;,\;(3)} \\{a^2} + \overrightarrow a \cdot \overrightarrow b + \overrightarrow a \cdot \overrightarrow c + \overrightarrow a \cdot \overrightarrow b + {b^2}\\ + \overrightarrow b \cdot \overrightarrow c + \overrightarrow a \cdot \overrightarrow b + \overrightarrow a \cdot \overrightarrow b + {c^2} = 0\\\\ \Rightarrow 2\overrightarrow a \cdot \overrightarrow b + 2\overrightarrow a \cdot \overrightarrow c + 2\overrightarrow b \cdot \overrightarrow c + {a^2} + {b^2} + {c^2} = 0\\\\ \Rightarrow 2(\overrightarrow a \cdot \overrightarrow b + \overrightarrow a \cdot \overrightarrow c + \overrightarrow b \cdot \overrightarrow c ) + 4 + 1 + 9 = 0\\\\ \Rightarrow 2(\overrightarrow a \cdot \overrightarrow b + \overrightarrow a \cdot \overrightarrow c + \overrightarrow b \cdot \overrightarrow c ) = - 14\\\\ \Rightarrow \overrightarrow a \cdot \overrightarrow b + \overrightarrow a \cdot \overrightarrow c + \overrightarrow b \cdot \overrightarrow c = - 7\end{array}\)

با فرض

\(\overrightarrow a = (1\;,\;1\;,\;1)\;,\\\overrightarrow b = (3\;,\;2\;,\; - 1)\;,\\\overrightarrow c = (5\;,\; - 2\;,\;1)\)

داریم :

\(\begin{array}{l}\overrightarrow a \cdot \overrightarrow b = 1 \times 3 + 1 \times 2 + 1 \times ( - 1) = 4\\\\\overrightarrow a \cdot \overrightarrow c = 1 \times 5 + 1 \times ( - 2) + 1 \times 1 = 4\\\\ \Rightarrow \overrightarrow a \cdot \overrightarrow b = \overrightarrow a \cdot \overrightarrow c \;,\;\overrightarrow b \ne \overrightarrow c \end{array}\)

\(\overrightarrow a = (1\;,\; - 3\;,\;4)\;,\\\overrightarrow b = (3\;,\; - 4\;,\;2)\;,\\\overrightarrow c = ( - 1\;,\;1\;,\;4)\)

پس:

\(\begin{array}{l}\overrightarrow d = \overrightarrow b + \overrightarrow c = (3\;,\; - 4\;,\;2) + ( - 1\;,\;1\;,\;4)\\ = (2\;,\; - 3\;,\;6)\\\\\overrightarrow a \cdot \overrightarrow d = 1 \times 2 + ( - 3) \times ( - 3) + 4 \times 6\\ = 2 + 9 + 24 = 35\\\\{d^2} = {2^2} + {( - 3)^2} + {6^2} = 4 + 9 + 36 = 49\\\\\overrightarrow {a'} = \frac{{\overrightarrow a \cdot \overrightarrow d }}{{{d^2}}}\overrightarrow d = \frac{{35}}{{49}}(2\;,\; - 3\;,\;6)\\\\ = \frac{5}{7}(2\;,\; - 3\;,\;6) = (\frac{{10}}{7}\;,\; - \frac{{15}}{7}\;,\;\frac{{30}}{7})\end{array}\)

\(\begin{array}{l}\overrightarrow a = (1\;,\; - 3\;,\;2)\;,\;\overrightarrow b = ( - 2\;,\;1\;,\;5)\\\\\overrightarrow c = \overrightarrow a \times \overrightarrow b = \left| {\begin{array}{*{20}{c}}{\overrightarrow i }&{\overrightarrow j }&{\overrightarrow k }\\1&{ - 3}&2\\{ - 2}&1&5\end{array}} \right| = 7\overrightarrow i - 9\overrightarrow j - 5\overrightarrow k \\\\ \Rightarrow \overrightarrow c = (7\;,\; - 9\;,\; - 5)\;,\;\overrightarrow c \bot \overrightarrow a \;,\;\overrightarrow c \bot \overrightarrow b \end{array}\)

\(\overrightarrow a = (1\;,\;2\;,\;3)\;,\\\overrightarrow b = (2\;,\;4\;,\;6)\;,\\\overrightarrow c = ( - 1\;,\; - 2\;,\; - 3)\)

داریم:

\(\begin{array}{l}\overrightarrow a \times \overrightarrow b = \left| {\begin{array}{*{20}{c}}{\overrightarrow i }&{\overrightarrow j }&{\overrightarrow k }\\1&2&3\\2&4&6\end{array}} \right| = \overrightarrow O \\\\\overrightarrow a \times \overrightarrow c = \left| {\begin{array}{*{20}{c}}{\overrightarrow i }&{\overrightarrow j }&{\overrightarrow k }\\1&2&3\\{ - 1}&{ - 2}&{ - 3}\end{array}} \right| = \overrightarrow O \\\\ \Rightarrow \overrightarrow a \times \overrightarrow b = \overrightarrow a \times \overrightarrow c \;,\;\overrightarrow b \ne \overrightarrow c \end{array}\)

\(\begin{array}{l}\left| {\overrightarrow a \times \overrightarrow b } \right| = \left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|\sin \theta \Rightarrow 72 = 3 \times 26 \times \sin \theta \\\\ \Rightarrow \sin \theta = \frac{{12}}{{13}}\\\\ \Rightarrow \cos \theta = \pm \sqrt {1 - {{\sin }^2}\theta } = \pm \sqrt {1 - \frac{{144}}{{169}}} \\\\ = \pm \sqrt {\frac{{25}}{{169}}} = \pm \frac{5}{{13}}\\\\\overrightarrow a \cdot \overrightarrow b = \left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|\cos \theta = 3 \times 26 \times ( \pm \frac{5}{{13}}) = \pm 30\end{array}\)

\(\begin{array}{l}A(3\;,\;5\;,\;7)\;,\;B(5\;,\;5\;,\;0)\;,\;C( - 4\;,\;0\;,\;4)\\\\\overrightarrow {AB} = (2\;,\;0\;,\; - 7)\;,\;\overrightarrow {AC} = ( - 7\;,\; - 5\;,\; - 3)\\\\\overrightarrow {AB} \times \overrightarrow {AC} = \left| {\begin{array}{*{20}{c}}{\overrightarrow i }&{\overrightarrow j }&{\overrightarrow k }\\2&0&{ - 7}\\{ - 7}&{ - 5}&{ - 3}\end{array}} \right| = \\ - 35\overrightarrow i + 55\overrightarrow j - 10\overrightarrow k \\\\ \Rightarrow \left| {\overrightarrow {AB} \times \overrightarrow {AC} } \right| = \sqrt {{{( - 35)}^2} + {{55}^2} + {{( - 10)}^2}} \\ = \sqrt {4350} \\\\{S_{A\mathop B\limits^\Delta C}} = \frac{1}{2}\left| {\overrightarrow {AB} \times \overrightarrow {AC} } \right| = \frac{{\sqrt {4350} }}{2}\end{array}\)