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Câu hỏi: Trong không gian $\text{Ox}yz$, cho $A\left( 0 ; -1 ; -1 \right)$, $B\left( -2 ; 1 ; 1 \right)$, $C\left( -1 ; 3 ; 0 \right)$, $D\left( 1 ; 1 ; 1 \right)$. Tính cosin của góc giữa hai đường thẳng $AB$ và $CD$ ?
A. $-\dfrac{\sqrt{3}}{3}$.
B. $-\dfrac{\sqrt{6}}{3}$.
C. $\dfrac{\sqrt{3}}{3}$.
D. $\dfrac{\sqrt{6}}{2}$.
A. $-\dfrac{\sqrt{3}}{3}$.
B. $-\dfrac{\sqrt{6}}{3}$.
C. $\dfrac{\sqrt{3}}{3}$.
D. $\dfrac{\sqrt{6}}{2}$.
$\overrightarrow{AB}=\left( -2 ; 2 ; 2 \right)$, $\overrightarrow{CD}=\left( 2 ; -2 ; 1 \right)$.
$cos\left( AB, CD \right)=\left| cos\left( \overrightarrow{AB}, \overrightarrow{CD} \right) \right|=\dfrac{\left| \overrightarrow{AB}.\overrightarrow{CD} \right|}{AB.CD}=\dfrac{6}{2\sqrt{3}.3}=\dfrac{1}{\sqrt{3}}$.
$cos\left( AB, CD \right)=\left| cos\left( \overrightarrow{AB}, \overrightarrow{CD} \right) \right|=\dfrac{\left| \overrightarrow{AB}.\overrightarrow{CD} \right|}{AB.CD}=\dfrac{6}{2\sqrt{3}.3}=\dfrac{1}{\sqrt{3}}$.
Đáp án C.