Đáp án:
\(\begin{array}{l}
\overrightarrow {AB} .\overrightarrow {AC} = 10\\
\overrightarrow {BC} .\overrightarrow {CD} = - 148\\
\overrightarrow {AD} .\overrightarrow {AC} = - 42\\
\overrightarrow {BD} .\overrightarrow {CD} = 78
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
\overrightarrow {AB} = \left( {1;1} \right);\,\,\overrightarrow {AC} = \left( {11; - 1} \right)\\
\Rightarrow \overrightarrow {AB} .\overrightarrow {AC} = 1.11 + 1.\left( { - 1} \right) = 11 - 1 = 10\\
\overrightarrow {BC} = \left( {10; - 2} \right);\,\,\overrightarrow {CD} = \left( { - 15; - 1} \right)\\
\Rightarrow \overrightarrow {BC} .\overrightarrow {CD} = 10.\left( { - 15} \right) + \left( { - 2} \right).\left( { - 1} \right) = - 148\\
\overrightarrow {AD} = \left( { - 4; - 2} \right);\,\,\overrightarrow {AC} = \left( {11; - 1} \right)\\
\Rightarrow \overrightarrow {AD} .\overrightarrow {AC} = - 4.11 + \left( { - 2} \right).\left( { - 1} \right) = - 42\\
\overrightarrow {BD} = \left( { - 5; - 3} \right);\,\,\overrightarrow {CD} = \left( { - 15; - 1} \right)\\
\Rightarrow \overrightarrow {BD} .\overrightarrow {CD} = - 5.\left( { - 15} \right) + \left( { - 3} \right).\left( { - 1} \right) = 78
\end{array}\)
