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It is a vector question.. Find the angle between the lines (x-1)/2 = 1-y = 2z and x = y = 3z.
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the lines are skew. \[\frac{x-1}{2}=\frac{1-y}{1}=\frac{2z}{1}\] or \[\frac{x-1}{2}=\frac{y-1}{-1}=\frac{z-0}{1/2}\] the vector parallel to the line is \[<2,-1,(1/2)>\] similarly, the vector parallel to the line \[x=y=3z\] is \[<1,1,(1/3)>\] find the angle between two vectors u and v by \[\Large \cos \theta = \frac{u\cdot v}{|u||v|}\]
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