# Plane and span Vector

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?

• There's no error. There's an infinity of pairs of spanning vectors. Oct 4, 2019 at 16:57
• 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?
– jl00
Oct 4, 2019 at 16:57
• jl00, it is [-4 0 1 ] and [2 1 0] Oct 4, 2019 at 17:41
• Bernard thanks very much Oct 4, 2019 at 20:17

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.