# Given the set W={(x ,y,z):ax +by +cz=0 a,b,c real } find a basis of W ?

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### 1 Answer

You can notice that the equation `ax + by + cz = 0` represents the equation of a plane and the plane is determined if the coefficients `a,b,c != 0` .

Considering `a!=0` , you may solve the equation of the plane, such that:

`ax = -by - cz => x = (-b/a)y - (c/a)z`

You need to consider the vectors `(x,y,z) = ((-b/a)y - (c/a)z, y,z)` and you need to write these vectors as the following linear combination, such that:

`(x,y,z) = y(-b/a,1,0) + z(-c/a,0,1)`

If the vectors `(-b/a,1,0)` and `(-c/a,0,1)` are linearly independent, then they form a basis of W, hence, you need to check if the vectors are linearly independent solving the following homogeneous system of equations, such that:

`barv_1(-b/a,1,0) + barv_2(-c/a,0,1) = (0,0,0)`

`{((-b/a)barv_1 - (c/a)barv_2 = 0),(barv_1 = 0),(barv_2 = 0):} => barv_1 = barv_2 = 0`

Since the solution of the system is the trivial solution, the vectors `(-b/a,1,0)` and `(-c/a,0,1)` are linearly independent.

**Since the vectors `(-b/a,1,0)` and `(-c/a,0,1)` are linearly independent, they form a basis of W.**

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