cho A(1,-2)B(3,0)C(-1,4)D2,-2) a,tính AB;CD b,tính vectơ AB, vectơ CD c, tính góc giữa vectơ AB và vectơ CD 09/11/2021 Bởi Everleigh cho A(1,-2)B(3,0)C(-1,4)D2,-2) a,tính AB;CD b,tính vectơ AB, vectơ CD c, tính góc giữa vectơ AB và vectơ CD
`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’` Bình luận
Đáp án: $\begin{array}{l}A\left( {1; – 2} \right);B\left( {3;0} \right);C\left( { – 1;4} \right);D\left( {2; – 2} \right)\\a)AB = \sqrt {{{\left( {3 – 1} \right)}^2} + {{\left( {0 + 2} \right)}^2}} = 2\sqrt 2 \\CD = \sqrt {{{\left( {2 + 1} \right)}^2} + {{\left( { – 2 – 4} \right)}^2}} = 3\sqrt 5 \\b)\overrightarrow {AB} = \left( {3 – 1;0 + 2} \right) = \left( {2;2} \right)\\\overrightarrow {CD} = \left( {2 + 1; – 2 – 4} \right) = \left( {3; – 6} \right)\\c)cos\left( {\overrightarrow {AB} ;\overrightarrow {CD} } \right) = \dfrac{{\overrightarrow {AB} .\overrightarrow {CD} }}{{AB.AC}}\\ = \dfrac{{2.3 + 2.\left( { – 6} \right)}}{{2\sqrt 2 .3\sqrt 5 }} = \dfrac{{ – 6}}{{6\sqrt {10} }} = \dfrac{{ – 1}}{{\sqrt {10} }}\\ \Rightarrow \left( {\overrightarrow {AB} ;\overrightarrow {CD} } \right) = {108^0}\end{array}$ Bình luận
`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’`
Đáp án:
$\begin{array}{l}
A\left( {1; – 2} \right);B\left( {3;0} \right);C\left( { – 1;4} \right);D\left( {2; – 2} \right)\\
a)AB = \sqrt {{{\left( {3 – 1} \right)}^2} + {{\left( {0 + 2} \right)}^2}} = 2\sqrt 2 \\
CD = \sqrt {{{\left( {2 + 1} \right)}^2} + {{\left( { – 2 – 4} \right)}^2}} = 3\sqrt 5 \\
b)\overrightarrow {AB} = \left( {3 – 1;0 + 2} \right) = \left( {2;2} \right)\\
\overrightarrow {CD} = \left( {2 + 1; – 2 – 4} \right) = \left( {3; – 6} \right)\\
c)cos\left( {\overrightarrow {AB} ;\overrightarrow {CD} } \right) = \dfrac{{\overrightarrow {AB} .\overrightarrow {CD} }}{{AB.AC}}\\
= \dfrac{{2.3 + 2.\left( { – 6} \right)}}{{2\sqrt 2 .3\sqrt 5 }} = \dfrac{{ – 6}}{{6\sqrt {10} }} = \dfrac{{ – 1}}{{\sqrt {10} }}\\
\Rightarrow \left( {\overrightarrow {AB} ;\overrightarrow {CD} } \right) = {108^0}
\end{array}$