# 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 $\langle 2,1,1\rangle$.

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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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? –  Sigur Dec 1 '12 at 23:39
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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 –  AAbc Dec 2 '12 at 0:22

## 2 Answers

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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