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Mathematics 18 Online
OpenStudy (anonymous):

the two lines x=ay+b, z= cy+d and x=a'y+b', z=c'y+d' are perpendicular to each other, if: (a) aa'+cc'=1 (b) (a/a') + (c/c') =-1 (c) (a/a')+(c/c')=1 (d) aa'+cc'=-1

OpenStudy (anonymous):

the answer would be d since we can rewrite the first two lines as y=(x-b)/a y=(z-d)/c therefore ((x-b)/a)= ((z-d)/c) which gives us <x,1,c> similar we write the second two lines is that for giving us <x',1,z'> and we know if they are perpendicular the dot product gives us 0 thus xx'+1+zz'=0

OpenStudy (anonymous):

ok how did you get <x,1,c>??

OpenStudy (anonymous):

i mixed up my signs halfway through it should be <a,1,c> and <a',1,c'> and aa'+1+cc'=0

OpenStudy (anonymous):

ohk that works.. thanx bro!

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