# Finding a basis for the plane with the equation

I know the conditions of being a basis. The vectors in set should be linearly independent and they should span the vector space.

So while finding a basis for the equation $y = z$, it's easy to see that the $x$ is free variable and if we call it as $x = s$, $y$ and $z$ becomes $t$ for example. And the basis are $(1,0,0)$ and $(0,1,1)$

However, i could not apply the same logic to $x - 2y + 5z =0$ Is there any free variable in this equation? How can I find the basis?

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As in your first example, you have two (not one) free variable. You let $x= s$, $y = t$ in the first example and $z = t$ follow from the equation. In this example you must take $x$ but you can choose between $y$ and $z$ to be free for you.

In the second example you can take any two variables as free. Let's take $x = s$, $y = t$. The equation $x-2y+5z = 0$ gives $$z = -\frac 15 x + \frac 25y = \frac 15(2t -s)$$ So a basis is given by $(1,0, -\frac 15)^\top$, $(0,1,\frac 25)^\top$.

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Oh thanks but the given basis is not your answer. – Yigit Can Dec 5 '12 at 21:22
Which given basis? – martini Dec 5 '12 at 21:23
i mean in the key – Yigit Can Dec 5 '12 at 21:23
What key? I don't unterstand? – martini Dec 5 '12 at 21:24
@martini The OP probably meant his/her solution manual. – user1551 Dec 6 '12 at 4:02

When you have the equation of a plane as $ax+by+cz=0$, the vector $(a,b,c)$ is perpendicular to the plane. So you simply need to find one vector contained in the plane by inspection, e.g. $(2,1,0)$, then get the second one as the cross product of the first vector with the perpendicular vector, i.e. $(2,1,0) \times (1, -2, 5) = (5, -10, -5)$.

So the pair of vectors $(2,1,0)$ and $(1, -2,-1)$ form not only a basis, but an orthogonal one, of your plane.

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