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Please help me work it out step by step :(

alright so you will have to use the cross product of the two normal vectors

I did and I got -12i-4j-20k

hm...how'd u get yours?

wait...i got it

practice makes perfect, good job so far cherio ;)

haha..gotta love calc =D..and ya i got -12i-4j-20k after i set it up right

:P

so this is your direction vector i believe

now you need to to find a point that lie on this line

you can do this by taking your two planes and setting x,y, or z to zero and finding a point

okay so if x=0 I get y=2/3 and z=-5/3 , right?

now i dot it with any scalar of the direction vector and get
L=(x, y-(2/3), z+(5/3))* t(-12,-4,-20)

|dw:1359565267353:dw|

i just set x=0, so i didn't have to do substitution. other way works too

okay..not to try to explain the picture...

i get a common point of (-2,0,-5) :) but any common point will suffice

Yes! thank you