How to find a basis of a subspace

[jeffreylze]

Homework Statement

Find bases for the following subspaces of R^3

(a) The set of vectors lying in the plane 2x-y-z=0
(b) The set of vectors on the line x/2=y/3=z/4

Homework Equations

The Attempt at a Solution

For part (a) , i tried using this method - X=x , y=y and z = 2x -y , hence (x , y, 2x-y). Then x(1,0,2)+y(0,1,-1) . Hence the basis is {(1,0,2) , (0,1,-1)} But then, i tried put y = 2x-z, x=x and z=z and i found a completely different answer. Why is that so? an x = y/2 + z/2 , y=y and z=z gives a different answer too..


For part (b) , I have absolutely no idea where to start.

 

[mighty2000]

Please help!

For part (a), the method you used is correct. The reason you are getting different answers is because there are multiple ways to represent a vector in a plane. For example, if you have a plane in R^3, you can represent it as a linear combination of two vectors that are parallel to the plane and not parallel to each other. In your case, (1,0,2) and (0,1,-1) are both parallel to the plane 2x-y-z=0 and are not parallel to each other, so they form a basis for the subspace.

For part (b), think about the equation x/2=y/3=z/4. This means that for any value of x, y and z are determined. So, let's say we choose x=2. Then y=3 and z=4. This gives us the vector (2,3,4). Now, we can write any other vector on this line as a scalar multiple of (2,3,4). So, the basis for this subspace is {(2,3,4)}.

I hope this helps! Let me know if you have any further questions.