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Câu hỏi: Cho hình lăng trụ tam giác đều $ABC.{A}'{B}'{C}'$ có $AB=a$ và $A{A}'=\sqrt{2} a$. Góc giữa hai đường thẳng $A{B}'$ và $B{C}'$ bằng
A. $60{}^\circ $.
B. $45{}^\circ $.
C. $90{}^\circ $.
D. $30{}^\circ $.
Ta có $\overrightarrow{A{B}'}.\overrightarrow{B{C}'}=\left( \overrightarrow{AB}+\overrightarrow{B{B}'} \right)\left( \overrightarrow{BC}+\overrightarrow{C{C}'} \right)$ $=\overrightarrow{AB}.\overrightarrow{BC}+\overrightarrow{AB}.\overrightarrow{C{C}'}+\overrightarrow{B{B}'}.\overrightarrow{BC}+\overrightarrow{B{B}'}.\overrightarrow{C{C}'}$
$=\overrightarrow{AB}.\overrightarrow{BC}+\overrightarrow{AB}.\overrightarrow{C{C}'}+\overrightarrow{B{B}'}.\overrightarrow{BC}+\overrightarrow{B{B}'}.\overrightarrow{C{C}'}$ $=-\dfrac{{{a}^{2}}}{2}+0+0+2{{a}^{2}}=\dfrac{3{{a}^{2}}}{2}$.
Suy ra $\cos \left( \overrightarrow{A{B}'},\overrightarrow{B{C}'} \right)=\dfrac{\overrightarrow{A{B}'}.\overrightarrow{B{C}'}}{\left| \overrightarrow{A{B}'} \right|.\left| \overrightarrow{B{C}'} \right|}$ $=\dfrac{\dfrac{3{{a}^{2}}}{2}}{a\sqrt{3}.a\sqrt{3}}=\dfrac{1}{2}\Rightarrow \widehat{\left( A{B}',B{C}' \right)}=60{}^\circ $.
B. $45{}^\circ $.
C. $90{}^\circ $.
D. $30{}^\circ $.
$=\overrightarrow{AB}.\overrightarrow{BC}+\overrightarrow{AB}.\overrightarrow{C{C}'}+\overrightarrow{B{B}'}.\overrightarrow{BC}+\overrightarrow{B{B}'}.\overrightarrow{C{C}'}$ $=-\dfrac{{{a}^{2}}}{2}+0+0+2{{a}^{2}}=\dfrac{3{{a}^{2}}}{2}$.
Suy ra $\cos \left( \overrightarrow{A{B}'},\overrightarrow{B{C}'} \right)=\dfrac{\overrightarrow{A{B}'}.\overrightarrow{B{C}'}}{\left| \overrightarrow{A{B}'} \right|.\left| \overrightarrow{B{C}'} \right|}$ $=\dfrac{\dfrac{3{{a}^{2}}}{2}}{a\sqrt{3}.a\sqrt{3}}=\dfrac{1}{2}\Rightarrow \widehat{\left( A{B}',B{C}' \right)}=60{}^\circ $.
Đáp án A.