sorry guys,but im writing exams tomorrow and just cant figure this out:

how do you calculate the distance from a plane to a line?

my plane is: (2/2/1) x X = 15 , X being a vector here

my line is: X= (2/3/2) + u x (1/-2/2) (u is also a vector)

i know that the distance is 1, but i cant figure out how to get there.. any help?
This is the MyImmortal law of math: If the class has the word math, or is in any way related to math, it shall be dropped, and if solution 1 can not happen it shall be nap time.

That's all the math help anybody needs. Glad I could help
If you know 'u' it's easy: You just replace the 'X' in the plane formula with the formula for 'X' (X=(2/3/2) + u x (1/-2/2)) so you'll get:

(2/2/1) × (2/3/2) + u × (1/-2/2) = 15

And that should be easy to solve.
(<X.X)O=('.'Q)

I'm the motherflippin'
If you know 'u' it's easy: You just replace the 'X' in the plane formula with the formula for 'X' (X=(2/3/2) + u x (1/-2/2)) so you'll get:

(2/2/1) × (2/3/2) + u × (1/-2/2) = 15

And that should be easy to solve.

i kindof have the feeling that you have no idea what im talking about, but thanks for trying anyways.
Quote by MyImmortal
This is the MyImmortal law of math: If the class has the word math, or is in any way related to math, it shall be dropped, and if solution 1 can not happen it shall be nap time.

That's all the math help anybody needs. Glad I could help

+1
There is no X

Or you can just devide it by zero.
Maybe you should draw a triangle, and use Pythagoras, like this:

Say that C is the position of the plane, and B is the end of the line (and the line goes on to the right, if you know what I mean), and you know the hight of the plane, maybe you could do something with that.