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anonymous
 3 years ago
Find the equation of the line of intersection of the planes with the equations 3x+4y+z=1 and 5x3z=5
anonymous
 3 years ago
Find the equation of the line of intersection of the planes with the equations 3x+4y+z=1 and 5x3z=5

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Please help me work it out step by step :(

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0alright so you will have to use the cross product of the two normal vectors

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I did and I got 12i4j20k

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0hm...how'd u get yours?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0practice makes perfect, good job so far cherio ;)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0haha..gotta love calc =D..and ya i got 12i4j20k after i set it up right

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so this is your direction vector i believe

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0now you need to to find a point that lie on this line

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you can do this by taking your two planes and setting x,y, or z to zero and finding a point

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0okay so if x=0 I get y=2/3 and z=5/3 , right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes, 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
 3 years ago
Best ResponseYou've already chosen the best response.0so 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 (x0, y(2/3), z+(5/3)) as that vector

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0now 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
 3 years ago
Best ResponseYou've already chosen the best response.0Any reason why you do (x0,y(2/3),z+(5/3)? and you can't simplify that any more so that would be the answer?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0well 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
 3 years ago
Best ResponseYou've already chosen the best response.0in 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
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1359565267353:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i just set x=0, so i didn't have to do substitution. other way works too

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0okay..not to try to explain the picture...

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0i get a common point of (2,0,5) :) but any common point will suffice

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0well 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
 3 years ago
Best ResponseYou've already chosen the best response.0crossing 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
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