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:
No graph is necessary. In fact, "a positive octant" is unnecessary.  
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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algebra, linear, planes 
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