A(0;1;2)vàd1:x/2=y−1/1=z+1/−1,d2:x=1+t;y=−1−2t;z=2+t. Tìm M∈d1,N∈d2 sao cho A,M,N thẳng hàng

Question

Cho \(A\left(0\, ;\, 1\, ;\, 2\right)và d_{1} :\frac{x}{2} =\frac{y-1}{1} =\frac{z+1}{-1} ,d_{2} :\left\{\begin{array}{l} {x=1+t} \\ {y=-1-2t} \\ {z=2+t} \end{array}\right. .\) Tìm tọa độ \(M\in d_{1} ,N\in d_{2} \) sao cho A,M,N thẳng hàng .
 

Answer 1

Do \(\left\{\begin{array}{l} {M\in d_{1} \Rightarrow M\left(2t_{1} ;1+t_{1} ;-1-t_{1} \right)} \\ {N\in d_{2} \Rightarrow N\left(1+t_{2} ;-1-2t_{2} ;2+t_{2} \right)} \end{array}\right. .\)

Ta có \(\overrightarrow{AM}=\left(2t_{1} ;t_{1} ;-3-t_{1} \right),\overrightarrow{AN}=\left(1+t_{2} ;-2-2t_{2} ;t_{2} \right). \)

Ba điểm A,M,N thẳng hàng \(\Leftrightarrow \overrightarrow{AM}=k\overrightarrow{AN}\Leftrightarrow \left\{\begin{array}{l} {2t_{1} =k\left(1+t_{2} \right)} \\ {t_{1} =k\left(-2-2t_{2} \right)} \\ {-3-t_{1} =kt_{2} } \end{array}\right. \Leftrightarrow \left\{\begin{array}{l} {t_{1} =0} \\ {t_{2} =-1} \\ {k=3} \end{array}\right. .\)

Vậy \(M\left(0;1;-1\right),N\left(0;1;1\right).\)