How Do You Find Non-Parallel Vectors in a Plane Defined by an Equation?

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SUMMARY

The discussion focuses on finding non-parallel vectors in a plane defined by the equation x + 3y + 2z = 0. Two specific vectors are derived: (1, 0, -1/2) and (0, 1, -3/2), which are confirmed to be independent and span the plane. The conversation also touches on the implications of the equation x + 3y + 2z ≠ 0, suggesting that such a condition would represent a different plane. The methodology involves selecting values for x and y to compute corresponding z values.

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Cankur
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Hello!

I was just thinking, say, I have a plane with the equation x + 3y + 2z = 0. How would I go about finding two vectors in that plane (that are not parallell)?

Sidenote: I pretty much get the intuition of how a plane with the given equation would look but what would happen if x + 3y + 2z ≠ 0? How would that plane look?

Thanks in advance!
 
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pick values for x and y, and use your equation to calculate z:
If x= 1, y= 0, then 1+ 3(0)+ 2z= 0 so z= -1/2. A point in that plane is (1, 0, -1/2) so the vector from (0, 0, 0) to (1, 0, -1/2), i- (1/2)k, is a vector in that plane.

If x= 0, y= 1, then 0+ 3(1)+ 2z= 0 so z= -3/2. A point in that plane is (0, 1, -3/2) so the vector from (0, 0, 0) to (0, 1, -3/2), j-(3/2)k, is another vector in that plane.

Of course, since those two vectors are independent, they span the plane.
 

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