I need to find the point in the plane x + 2y + 3z = 6 that is nearest the origin.
So the parametric equations of the line passing through the origin and in the direction of this normal vector will be:
x = at
y = bt
z = ct
The point where this line cuts the plane will be the point that's closest to the origin:
$\displaystyle a(at) + b(bt) + c(ct) = d \Rightarrow t = \frac{d}{a^2 + b^2 + c^2}$.
Therefore the coordinates of the point in the plane closest to the origin are:
$\displaystyle x = at = \frac{ad}{a^2 + b^2 + c^2}$.
$\displaystyle y = bt = \frac{bd}{a^2 + b^2 + c^2}$.
$\displaystyle x = at = \frac{cd}{a^2 + b^2 + c^2}$.
And as a bonus, the answer to what the closest distance is equal to is ..........