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I have the following problem: the plane equation is given by $x_1-2x_2+4x_3=0$ I need to come up with the two vectors that spans the plane. So the normal from the equation can be written as $(1,-2,4)$ So $(1,-2,4)$ has to be equal to the cross product of two vectors:

$$1=a_2b_3-a_3b_2$$ $$-2=a_3b_1-a_1b_3$$$$ 4=a_1b_2-a_2b_1$$

So I take the random approach and set $b_1$ to $0$, so I obtain $\frac{b_3}{b_2}=\frac12$.So $b_3=1$, and $b_2=2$. If I replace these values in the system, I obtain $a_1=2$, $a_2=3$, $a_3=1$. So two vectors can be span ($[2,3,1]$ and $[0,2,1]$). However, the answer in the back of the book is different. Can someone explain where I make mistake?

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  • $\begingroup$ There's no error. There's an infinity of pairs of spanning vectors. $\endgroup$
    – Bernard
    Commented Oct 4, 2019 at 16:57
  • $\begingroup$ What are the answers in the back of book? Perhaps the spans are the same, but they wrote the vectors in the basis as some linear combination of your vectors? $\endgroup$
    – jl00
    Commented Oct 4, 2019 at 16:57
  • $\begingroup$ jl00, it is [-4 0 1 ] and [2 1 0] $\endgroup$ Commented Oct 4, 2019 at 17:41
  • $\begingroup$ Bernard thanks very much $\endgroup$ Commented Oct 4, 2019 at 20:17

1 Answer 1

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As noted in comments, there’s an infinite number of solutions to this problem. However, observe that for any nonzero vector $(a,b,c)$, the vectors $(0,c,-b)$, $(-c,0,a)$ and $(b,-a,0)$ are all orthogonal to $(a,b,c)$, and at least two of them are nonzero. (These vectors are the cross products of $(a,b,c)$ with the standard basis vectors.) Their negations are, of course, also orthogonal to $(a,b,c)$.

So, for your normal vector $(1,-2,4)$ We can immediately find three nonzero vectors orthogonal to it: $(0,4,2)$, $(-4,0,1)$ and $(-2,-1,0)$. The second of these and the negation of the last one give you the book’s answer.

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