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

• 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? Dec 1, 2012 at 23:39
The technique is to find the equation of the plane $ax+by+cz+d=0$, then use the formula of the distance
$$D=\frac{ax_0+by_0+cz_0+d}{\sqrt{a^2+b^2+c^2}}.$$
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.