Question

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

Answer 1

Please help me work it out step by step :(

Answer 2

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

Answer 3

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

Answer 4

hm...how'd u get yours?

Answer 5

wait...i got it

Answer 6

practice makes perfect, good job so far cherio ;)

Answer 7

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

Answer 8

:P

Answer 9

so this is your direction vector i believe

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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)

Answer 14

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

Answer 15

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)

Answer 16

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?

Answer 17

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.

Answer 18

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 ?