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KatClaire
Find the equation of the line of intersection of the planes with the equations -3x+4y+z=1 and 5x-3z=5
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?
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
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?
yes, so for the next part you have to make a vector with that point (this will be dotted with the direction vector from before)
so typically for a line you just select the (x,y,z) to be your point (if anything doesn't make sense let me know) so you get (x-0, y-(2/3), z+(5/3)) as that vector
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)
Any reason why you do (x-0,y-(2/3),z+(5/3)? and you can't simplify that any more so that would be the answer?
well if you want to write in parametric or (something that starts with a s...symmetric?) you can simplify it a bit more..but as far at why x,y,z...um..i'll try to draw a picture.
in case i missed it, since the second plane has no y value, wouldnt it be simper to make y=0 in the first to solve for x,z ?
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
well either way, I think I understand it more now! I was confused as to why the cross product was involved and what not. Thanks!!
crossing vectors created a new vector perp to both crossing 2 plane normals will get you a direction vector that is parallel to the line of insection