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May 22nd, 2015, 09:44 AM   #1
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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?
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May 22nd, 2015, 10:28 AM   #2
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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).
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May 22nd, 2015, 03:34 PM   #3
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Hello, Lekha!

Quote:
Sketch the plane on a positive octant where
Do the same for on the same graph.
What vector is perpendicular to these two planes?

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





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May 22nd, 2015, 03:43 PM   #4
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A minor quibble: it's a normal vector, not the normal vector. $\langle 2,2,2 \rangle$ is another.
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