Find the angle in radians between the planes -2x +z =1 and 3y+z =1.
Really need to know how to do this.

Question

Find the angle in radians between the planes -2x +z =1 and 3y+z =1.
Really need to know how to do this.

saifoo.khan · Answer

@satellite73

saifoo.khan · Answer

Nope, sorry.

anonymous · Answer

Yeah I looked up the answer and it didn't work :/

anonymous · Answer

Do you know how to go about this it makes no sense to me

anonymous · Answer

anonymous · Answer

Alright thanks a bunch!

anonymous · Answer

Here's a long shot...Since y doesn't occur in the first equation and x doesn't occur in the second one, does that mean that they are at 90° ?

anonymous · Answer

It has to be in radians and I have tried pi/2 and it doesnt seem to work :/

anonymous · Answer

anonymous · Answer

-2x +z =1 and 3y+z =1
n1 = -2,0,1
n2 = 0,3,1
n1.n2 = 1
cos(theta) = 1
theta = 0

anonymous · Answer

Oops, I forgot the denominator of the dot product...

anonymous · Answer

n1.n2 = 1/(sqrt(5)*sqrt(10)) = 1/(5sqrt(2))
arccos(1/(5sqrt(2)))

anonymous · Answer

You got it. You are my hero. Legit! Thanks a bunch!

anonymous · Answer

81.87 degrees?

anonymous · Answer

I just answer the arccos (1/(5sqrt(2))) and it was right haha

anonymous · Answer

I'm surprised it worked.  I've never done those before.

anonymous · Answer

Haha well I am amazed thanks a lot!

anonymous · Answer

Thanks for the fun problem, see you!