Find the equation of the line of intersection of the planes with the equations -3x+4y+z=1 and 5x-3z=5

- anonymous

- schrodinger

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- anonymous

Please help me work it out step by step :(

- anonymous

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

- anonymous

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

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## More answers

- anonymous

hm...how'd u get yours?

- anonymous

wait...i got it

- amistre64

practice makes perfect, good job so far cherio ;)

- anonymous

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

- anonymous

:P

- anonymous

so this is your direction vector i believe

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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)

- anonymous

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

- anonymous

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)

- anonymous

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?

- anonymous

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.

- amistre64

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 ?

- anonymous

|dw:1359565267353:dw|

- anonymous

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

- anonymous

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

- amistre64

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

- anonymous

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!!

- amistre64

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

- anonymous

Yes! thank you

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