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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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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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