Question

Đã trả lời bởi chuyên gia

Answer 1

Ta có: $\vec{M N} = \frac{1}{2} (\vec{A C} + \vec{B D}) \Rightarrow {|\vec{M N}|}^{2} = \frac{1}{4} ({\vec{A C}}^{2} + 2 \vec{A C} . \vec{B D} + {\vec{B D}}^{2})$

$ = \frac{1}{4}\left( {2{a^2} + 2\overrightarrow {AC} .\overrightarrow {BD} } \right).$

Mà: $\overrightarrow {AC} .\overrightarrow {BD} = \overrightarrow {AC} .\left( {\overrightarrow {AD} - \overrightarrow {AB} } \right) = \overrightarrow {AC} .\overrightarrow {AD} - \overrightarrow {AC} .\overrightarrow {AB} $

$ = \left| {\overrightarrow {AC} } \right|.\left| {\overrightarrow {AD} } \right|.\cos 60^\circ - \left| {\overrightarrow {AC} } \right|.\left| {\overrightarrow {AB} } \right|.\cos 60^\circ = 0.$

Suy ra ${\left| {\overrightarrow {MN} } \right|^2} = \frac{1}{4}.2{a^2} = \frac{{{a^2}}}{2}$$ \Rightarrow \left| {\overrightarrow {MN} } \right| = \frac{{a\sqrt 2 }}{2}.$

Ta có: $\vec{A C} . \vec{M N} = \frac{1}{2} \vec{A C} . (\vec{A C} + \vec{B D}) = \frac{1}{2} ({\vec{A C}}^{2} + \vec{A C} . \vec{B D}) = \frac{a^{2}}{2} .$

Khi đó, $\cos (\vec{A C}, \vec{M N}) = \frac{\vec{A C} . \vec{M N}}{|\vec{A C}| . |\vec{M N}|} = \frac{\frac{a^{2}}{2}}{a . \frac{a \sqrt{2}}{2}} = \frac{\sqrt{2}}{2} .$

Vậy $\cos \left( {\overrightarrow {AC} ,\overrightarrow {MN} } \right) = \frac{{\sqrt 2 }}{2}.$

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Answer 2