How can I find orthogonal vectors in $\mathbb{R}^3$ that lie in a plane created by the equation below:
$$x_1-2x_2+4x_3=0$$
Thanks,
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1$$\vec{n}=(1;-2;4)$$ is such a vector – Dr. Sonnhard Graubner Sep 29 '17 at 18:34
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or $$[-2;1;4]\times [0;2;1]$$ – Dr. Sonnhard Graubner Sep 29 '17 at 18:36
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@Dr.SonnhardGraubner $(1; -2; 4)$ is orthogonal to the plane. The OP wants orthogonal vectors that lie in the plane. – Sep 29 '17 at 18:40
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ok i misread this question – Dr. Sonnhard Graubner Sep 29 '17 at 18:41
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@Dr.SonnhardGraubner So did I, probably due to the misleading title... – Sep 29 '17 at 18:42
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A plane has the equation $$a x_1+b x_2+c x_3=d$$ The normal to this plane is $n=[a,b,c]$.
So you want two vectors perpendicular to each other, and perpendicular to $n$.
Pick any vector $v_0$ not parallel to $n$. Then $v_1=n\times v_0$ and $v_2=n\times v_1$ are the sought-after vectors. (see cross product)
Wouter
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From the equation of the plane, we find that the vector $\mathbf n=(1,-2,4)^T$ is normal to the plane. Every vector that’s orthogonal to $\mathbf n$ lies in the plane. Three obvious choices are $(2,1,0)^T$, $(4,0,-1)^T$ and $(0,4,2)^T$. (In general, given a non-zero vector $(a,b,c)^T$, at least two of $(0,c,-b)^T$, $(-c,0,a)^T$ and $(b,-a,0)^T$ are non-zero and orthogonal to the original vector.) Choose any one of these for your first vector and the cross product of $\mathbf n$ with the chosen vector for the other.
Essentially, you’ve being asked to find an orthogonal basis of $\mathbf R^3$ such that two of the basis vectors span the given plane. In $\mathbb R^3$ you can take advantage of cross products to do this. In higher-dimensional spaces, you’ll have to do something a little different. In this answer, Marc van Leeuwen describes an easy way to construct such a basis for a hyperplane in $\mathbb R^n$.
amd
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