A. \(30^\circ \)
B. \(60^\circ \)
C. \(45^\circ \)
D. \(120^\circ \)
GY:
Ta có: \(AB’ = BC’ = a\sqrt 3 \)
\(\overrightarrow {AB’} .\overrightarrow {BC’} = \left( {\overrightarrow {BB’} + \overrightarrow {A’B’} } \right).\left( {\overrightarrow {BB’} + \overrightarrow {B’C’} } \right)\)\( = B'{B^2} + \overrightarrow {A’B’} .\overrightarrow {B’C’} = 2{a^2} – \frac{1}{2}{a^2} = \frac{3}{2}{a^2}\).
\( \Rightarrow \cos \left( {\overrightarrow {AB’} ,\overrightarrow {BC’} } \right) = \frac{{\overrightarrow {AB’} .\overrightarrow {BC’} }}{{AB’.BC’}} = \frac{{\frac{3}{2}{a^2}}}{{3{a^2}}} = \frac{1}{2}\)
Vậy góc giữa \(AB’\) và \(BC’\) bằng \(60^\circ \).
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