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Find the minimal distance from the point $(8,−2,−6)$ to the plane $V$ in $\Bbb R^3$ spanned by $\langle -2,-2,2 \rangle$ and $\langle 2,1,1\rangle$.

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  • $\begingroup$ We need one point of the plane to know its position. Only 2 vectors is not enough to determine the plane. Are you supposing $V$ containing the origin? $\endgroup$
    – Sigur
    Dec 1, 2012 at 23:39
  • $\begingroup$ Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. Also, many find the use of imperative ("Prove", "Solve", etc.) to be rude when asking for help; please consider rewriting your post. $\endgroup$ Dec 1, 2012 at 23:46
  • $\begingroup$ hi this is linear algebra problem. and if my 2 vector are orthogonal then I could use projection to find the shortest point. and by using distance formula I can get the minimal distance. But my problem now is my vectors are not orthogonal $\endgroup$
    – AAbc
    Dec 2, 2012 at 0:22

2 Answers 2

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The technique is to find the equation of the plane $ ax+by+cz+d=0 $, then use the formula of the distance

$$ D=\frac{ax_0+by_0+cz_0+d}{\sqrt{a^2+b^2+c^2}}.$$

For more details see here.

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We know that in order for vectors to be orthogonal their dot product must equal to $0$. Since your vectors aren't orthogonal you can use Gram Schmidt process to orthogonalize the given vectors. Once you use that method, then you can use projections to find the minimum distance. If you need me to elaborate any further just ask.

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