Question

Determine whether the surfaces x^2+8x+y^2-9y+z^2-6z+65=0 and x^2+y^2=3x-2y intersect. do i use vectors?

Answer 1

I would try writing it in this form first (x-h)^2+(y-k)^2=r^2

Answer 2

both of them

Answer 3

\[x^2+8x+4^2+y^2-9y+(\frac{9}{2})^2+z^2-6z+3^2=-65+4^2+(\frac{9}{2})^2+3^2\]

Answer 4

i didn't realize that one was a sphere until i seen all the z's
oh i got cut off

Answer 5

oh shoot sorry, it's not +65, it's -85 at the end

Answer 6

oh no i didn't get cut off

Answer 7

is the 2nd equation still a sphere? becasue there aren't any z components

Answer 8

\[x^2+8x+4^2+y^2-9y+(\frac{9}{2})^2+z^2-6z+3^2=-85+4^2+(\frac{9}{2})^2+3^2\]

Answer 9

its a cirlce

Answer 10

is this correct?

Answer 11

ohh ok. yeah that makes sense

Answer 12

so i have =-85 not =85 right?

Answer 13

\[(x+4)^2+(y-\frac{9}{2})^2+(z-3)^2=-85+16+\frac{81}{4}+9\]

Answer 14

well it should be ....\[z^{2}\]-6x-85,
so it should be =85
does that make sense?

Answer 15

ok cool

Answer 16

\[(x+4)^2+(y-\frac{9}{2})^2+(z-3)^2=85+16+\frac{81}{4}+9\]
so that means we have this instead

Answer 17

yes

Answer 18

should i be setting the radii equal to eachother? or using parametrics?

Answer 19

\[(x+4)^2+(y-\frac{9}{2})^2+(z-3)^2=130.25\]

Answer 20

i haven't thought that far ahead

Answer 21

i'm still thinking as i go

Answer 22

\[x^2-3x+y^2+2y=0\]

Answer 23

so this is that second equation

Answer 24

yeah

Answer 25

\[x^2-3x+(\frac{3}{2})^2+y^2-2y+1^2=0+(\frac{3}{2})^2+1^2\]

Answer 26

\[(x-\frac{3}{2})^2+(y-1)^2=\frac{9}{4}+1\]

Answer 27

\[(x-\frac{3}{2})^2+(y-1)^2=3.25\]

Answer 28

so what to do next....thinking....

Answer 29

justdo the cros product of the vectors of the planes if it trunsout to be0 then they dont intersect

Answer 30

ohhh ok thanks. that makes a little more sense.

Answer 31

Show me how to find the vector of the plane. I want to see.

Answer 32

hmmm i thought i had it, but looking i;m not finding what i want in my text book

Answer 33

Do you know how to write the vectors?

Answer 34

yeah, but not from that equation. just from points or components given.

Answer 35

I think I was going the wrong way about it

Answer 36

earlier

Answer 37

oh, by completing the square?

Answer 38

write now i'm looking at pauls notes
and you were write we need to write the equations in parametric form

Answer 39

gahhhhhhh

Answer 40

lol right binary
to find how how to write the vectors we need to write the equation in parametric form?

Answer 41

well maybe we need the radius
and putting it in that form does give us the radius

Answer 42

so maybe it wasn't a total waste

Answer 43

take two points on the circle and find their vectors from center. do this for both the circles and go for crossproduct.

Answer 44

this will give the plane vector

Answer 45

if dcXio gives you some thing then it means the plane intersect but for circles it shoud be worked out.