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I'm being asked to complete the last two vectors to make $\vec{x}, \vec{y}, \vec{z}$ an orthogonal set:

$$\vec{x}=(1, 1, 1), \vec{y}=(a, b, 1), \vec{z}=(c, d, 1)$$

I thought I needed to use Graham-Schmidt, but it seems to lead me through a rabbit hole of arbitrary calculations that look wrong. Any help with this would be super appreciated.

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  • $\begingroup$ Instead, you would just need to solve the equations $x \cdot y = y \cdot z = z\cdot x = 0$. This may have more than one solution, though. $\endgroup$ Dec 14, 2017 at 6:31
  • $\begingroup$ I'm assuming it has a lot of answers. I'm not sure I know how to go about solving a system of equations with 4 variables though... $\endgroup$
    – S. Seki.
    Dec 14, 2017 at 6:34
  • $\begingroup$ $a+b=c+d = -1, $ and $ac=-bd$. Fix one of them, say $a=1$, and then $b=-2$, which gives you a set of two equations in two variables $c,d$. You could have chosen any other value instead of $1$, and you would have probably succeeded in finding $c,d$ uniquely. $\endgroup$ Dec 14, 2017 at 6:37
  • $\begingroup$ Thanks! I will definitely give this a try. $\endgroup$
    – S. Seki.
    Dec 14, 2017 at 6:42
  • $\begingroup$ @S.Seki. Why do you want to find "all" possible values when your question doesn't ask you to do that? Only one solution for $a,b,c,d$ is enough to make the given set orthogonal. $\endgroup$ Dec 14, 2017 at 6:54

2 Answers 2

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Just pick $a$ and $b$ such that $a+b+1=0$. Then vectors $x$ and $y$ are orthogonal. Consider their cross product; unless its last entry is zero one can divide by it and get $z$ orthogonal to $x$ and $y$ with last entry $1$.

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In our case, orthogonality is translated to the following system of equations: $$(1,1,1) \cdot (a,b,1) = a + b + 1 = 0\implies a=-(1+b)$$ $$(1,1,1) \cdot (c,d,1) = c + d + 1 = 0\implies c=-(1+d)$$ $$(a,b,1) \cdot (c,d,1) = ac+bd + 1 =0\implies b+d+2bd+2=0$$

Your problem is not asking you to find all possible values, but only one value to make the set orthogonal. To find one possible value, assume $b=-d$ and you will get:

$$-2b^2+2=0 \implies b^2=1 \implies b=\pm 1$$

So, you find two solutions for $(a,b,c,d)$: $\{(-2,1,0,-1), (0,-1,-2,1)\}$

Picking the first one, gives us that $\{(1,1,1),(-2,1,1),(0,-1,1)\}$ is an orthogonal set and solves your problem.

However, you could go further than this. Suppose that $b = \lambda d$ where $\lambda$ is a variable to obtain:

$$2\lambda d^2 + (\lambda + 1)d + 2 = 0$$

Therefore,

$$d = -\frac{\lambda + 1 \pm\sqrt{\lambda^2 - 14\lambda + 1}}{4 \lambda}$$

which is valid when $\lambda^2 - 14 \lambda + 1 > 0$

Now you can calculate $b$ and then $a,c$ respectively. Therefore, all possible $4$-tuples $a,b,c,d$ satisfying your conditions are parametrized by $\lambda$.

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  • $\begingroup$ I see what you mean, thanks so much for your help! $\endgroup$
    – S. Seki.
    Dec 16, 2017 at 5:17
  • $\begingroup$ @S.Seki.You're welcome. $\endgroup$ Dec 16, 2017 at 9:47

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