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 May 22nd, 2015, 09:44 AM #1 Newbie   Joined: May 2015 From: india Posts: 1 Thanks: 0 Linear algebra Sketch the plane x+y+z=1 on a positive octant where x=>0,y=>0 and z>=0. Do the same for x+y+z=2 on th same plane. What vector is perpendicular to these two planes?
 May 22nd, 2015, 10:28 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 Have you done what this problem suggests- graph the two planes? In any case, there exist an infinite number of vectors that are perpendicular to a given plane- though they are all multiples of the same vector. One way to find such a perpendicular is to take two vectors in the plane and take the cross product of them. One plane is x+ y+ z= 1. It is easy to see that (1, 0, 0), (0, 1, 0), and (0, 0, 1) are points in the plane so the vectors <1, -1, 0>, which is from (0, 1, 0) to (1, 0, 0) and <1, 0, -1> which is the vector from (0, 0, 1) to (1, 0, 0).
May 22nd, 2015, 03:34 PM   #3
Math Team

Joined: Dec 2006
From: Lexington, MA

Posts: 3,267
Thanks: 408

Hello, Lekha!

Quote:
 Sketch the plane $x\,+\,y\,+\,z\:=\:1$ on a positive octant where $x,y,z \ge 0.$ Do the same for $x\,+\,y\,+\,z\:=\:2$ on the same graph. What vector is perpendicular to these two planes?

No graph is necessary.
In fact, "a positive octant" is unnecessary.

$\text{The normal vector to the plane }ax\,+\,by\,+\,cz \:=\:d \, \text{ is: }\,\vec{n} \:=\:\langle a,\,b,\,c\rangle$

$\text{Therefore, the normal vector to }\,\begin{Bmatrix}x\,+\,y\,+\,z \:=\:1 \\ \\ x\,+\,y\,+\,z \:=\:2 \end{Bmatrix}\, \text{ is: }\,\vec{n} \:=\:\langle 1,\,1,\,1\rangle.$

 May 22nd, 2015, 03:43 PM #4 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,634 Thanks: 2620 Math Focus: Mainly analysis and algebra A minor quibble: it's a normal vector, not the normal vector. $\langle 2,2,2 \rangle$ is another.

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