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find the angle between
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lines (x-2)/1=(1-y)/3=Z-3/1 and x/2=y+3/-1=z-1/2
\[(x-2)/1=(1-y)/3=(z-3)/1\] and \[\frac{ x }{ 2 }=\frac{ y+3 }{ -1 }=\frac{ z-1 }{ 2 }\]
O_O It's SHinchan !
Ahh I don't remember how to do this? D: Do we want to find the vectors normal to the surfaces? Oh it's a line... +_+ Hmmm
\[a.b=|a||b| \cos \theta\]
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So do we want to rewrite the lines in parametric form and mess with them from there?
i tried to find the normal vector of them which is n1<1,3,1> and n2<2,-1,2> and used the formula but didn't work.
I think the normal to the first line is <1,-3,1> isn't it? :o
\[1) x=1+2t, y=1-3t, z=3+t\]
\[1) x=\color{red}{2+t}, y=1-3t, z=3+t\]
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dang it, i messed up "<1,-3,1>" -3 that the loophole
thx @zepdrix
i got 45.3
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