Đáp án:

$\begin{array}{l}
A\left( {1; - 2} \right);B\left( {3;0} \right);C\left( { - 1;4} \right);D\left( {2; - 2} \right)\\
a)AB = \sqrt {{{\left( {3 - 1} \right)}^2} + {{\left( {0 + 2} \right)}^2}}  = 2\sqrt 2 \\
CD = \sqrt {{{\left( {2 + 1} \right)}^2} + {{\left( { - 2 - 4} \right)}^2}}  = 3\sqrt 5 \\
b)\overrightarrow {AB}  = \left( {3 - 1;0 + 2} \right) = \left( {2;2} \right)\\
\overrightarrow {CD}  = \left( {2 + 1; - 2 - 4} \right) = \left( {3; - 6} \right)\\
c)cos\left( {\overrightarrow {AB} ;\overrightarrow {CD} } \right) = \dfrac{{\overrightarrow {AB} .\overrightarrow {CD} }}{{AB.AC}}\\
 = \dfrac{{2.3 + 2.\left( { - 6} \right)}}{{2\sqrt 2 .3\sqrt 5 }} = \dfrac{{ - 6}}{{6\sqrt {10} }} = \dfrac{{ - 1}}{{\sqrt {10} }}\\
 \Rightarrow \left( {\overrightarrow {AB} ;\overrightarrow {CD} } \right) = {108^0}
\end{array}$

`A(1,-2);B(3,0);C(-1,4);D(2,-2)`

`a)` Ta có:

`AB=\sqrt{(3-1)^2+(0+2)^2}=\sqrt{8}=2\sqrt{2}`

`CD=\sqrt{(2+1)^2+(-2-4)^2}=\sqrt{45}=3\sqrt{5}`

`b)` `\vec{AB}=(x_B-x_A;y_B-y_A)=(3-1;0+2)`

`\vec{AB}=(2;2)`

`\vec{CD}=(x_D-x_C;y_D-y_C)=(2+1;-2-4)`

`\vec{CD}=(3;-6)`

`c)` Ta có:

`\vec{AB}.\vec{CD}=|\vec{AB}.|\vec{CD}|.cos(\vec{AB};\vec{CD})`

`=>cos(\vec{AB};\vec{CD})={\vec{AB}.\vec{CD}}/{|\vec{AB}|.|\vec{CD}|}`

`={2.3+2.(-6)}/{2\sqrt{2}.3\sqrt{5}}={-6}/{6\sqrt{10}}={-1}/{\sqrt{10}}`

`=>(\vec{AB};\vec{CD})≈108°26'`

Vậy góc giữa hai vecto `\vec{AB}` và `\vec{CD}` khoảng `108°26'`