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 February 3rd, 2011, 07:52 AM #1 Newbie   Joined: Feb 2011 Posts: 3 Thanks: 0 Angle between two planes, urgent! The task is sounds as following: A plane (alpha) goes has the equation 2x+y+2z-1=0 Another plane (beta) is parallel with the vectors [1,1,2] and [2,-1,5]. Find the angle between the planes. At the moment, I have really no idea how I should solve this, so if someone could point me in the right direction, I would be very happy. Thanks in advance, Chris.
 February 3rd, 2011, 09:14 AM #2 Newbie   Joined: Feb 2011 Posts: 3 Thanks: 0 Re: Angle between two planes, urgent! Allright, well I solved this myself, so never mind!
February 3rd, 2011, 10:02 AM   #3
Math Team

Joined: Dec 2006
From: Lexington, MA

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Re: Angle between two planes, urgent!

Hello, Skovisen!

You solved it . . . Good for you!

Quote:
 $\text{A plane }(\alpha)\text{ has the equation: }\:2x\,+\,y\,+\,2z-1\:=\:0$ $\text{Another plane }(\beta)\text{ is parallel to the vectors }\langle1,\,1,\,2\rangle\text{ and }\langle2,\,-1,\,5\rangle.$ $\text{Find the angle between the planes.}$

Here's my approach . . .

$\text{Plane }\alpha\text{ has normal vector: }\:\vec u \:=\:\langle 2,\,1,\,2\rangle$

$\text{The normal vector of plane }\beta\text{ is perpendicular to the two vectors:}$

[color=beige]. . [/color]$\vec v \;=\;\begin{vmatrix} i &j=&k \\ 1=&1=&2 \\ \\ 2=&-1=&5 \end{vmatrix} \;=\; \langle 7,\,-1,\,-3\rangle$

$\text{The angle between the two planes is the angle between the two normal vectors.}$

[color=beige]. . [/color]$\cos\theta \;=\;\frac{\vec u\,\cdot\,\vec v}{|\vec u||\vec v|} \;=\; \frac{\langle2,\,1,\,2\rangle\,\cdot\,\langle7,\,-1,\,-3\rangle}{\sqrt{4 + 1 + 4}\,\sqrt{49+1+9}} \;=\;\frac{14\,-\,1\,-\,6}{\sqrt{9}\,\sqrt{59}} \;=\; \;=\;\frac{7}{3\sqrt{59}} \;=\;0.303774126$

$\text{Therefore: }\;\theta \;=\;72.31557243^o \;\approx\;72.3^o$

 February 3rd, 2011, 11:54 AM #4 Newbie   Joined: Feb 2011 Posts: 3 Thanks: 0 Re: Angle between two planes, urgent! Exactly what I did myself! Just took me some time to figure it out, but turned out to be not as hard as I first thought.

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