Question

Two planes have equations:
3x-y+2z=9 & x+y-4z=-1
Find a vector equation of the line of intersection of the planes.
- Not really sure how to do this one...

Answer 1

we know the normal vectors of each plane
the first plane has normal (3, -1, 2)
the second plane has normal (1,1 -4 )

Answer 2

Ok I understand that, but what is the = 9 and = -1 mean? I'm not to sure.

Answer 3

you can think of it as the height (we translate the plane through the origin)

Answer 4

Uh what exactly are you trying to show there?

Answer 5

i thought you were asking about the constant on the right side

Answer 6

vector calculus.... no it's asking for an equation that intersects... I haven't done calc iii in 2 years .... so I forgot the formula ^^

Answer 7

I think I get the height part, I just really hate vectors...

Answer 8

here is an example http://www.netcomuk.co.uk/~jenolive/vect18d.html

Answer 9

actually vector calculus is much easier than calculus i and ii XD. It's just that it's been a while...but it's not too hard to refresh my memory on calc iii

Answer 10

so alls we need to do is
(3, -1, 2) crossproduct (1,1 -4 )

Answer 11

AH! It's starting to come back

Answer 12

The cross product of (2, -5, 3) and (3, 4, -3) is (3, 15, 23).
How exactly does that work?
I don't think I ever learnt about the cross product...

Answer 13

so we either cross product u v or cross product u v w

Answer 14

depending on whether or not we are in R^2 or R^3

Answer 15

huh? I don't really follow what you're saying but this looks correct:
https://courses.eas.ualberta.ca/eas421/diagramspublic/equations/vectorcrossproduct.gif

Answer 16

http://tutorial.math.lamar.edu/Classes/CalcII/CrossProduct.aspx just look at one of the examples on here

Answer 17

example 1 ... has what you're looking for

Answer 18

It's coming back a bit, now I learnt it in a different format so yeah -_-.

So for the cross product of (3, -1, 2) x (1, 1, -4) I got (2, 14, 4)

Answer 19

ok imagine we have two planes that intersect