I need to find the point in the plane x + 2y + 3z = 6 that is nearest the origin.

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- Feb 12th 2008, 02:25 PMchaddyclosest point
I need to find the point in the plane x + 2y + 3z = 6 that is nearest the origin.

- Feb 12th 2008, 02:32 PMecMathGeek
- Feb 12th 2008, 02:34 PMchaddy
How would I find the normal vector?

- Feb 14th 2008, 07:34 PMchaddy
Does anyone know?

- Feb 14th 2008, 08:00 PMmr fantastic
- Feb 14th 2008, 08:06 PMmr fantastic
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 .......... - Feb 15th 2008, 12:10 AMchaddy
$\displaystyle

x = \frac{6}{14}

$

$\displaystyle

y = \frac{12}{ 14}

$

$\displaystyle

z = \frac{18}{ 14}

$

So the point closest to the origin would be < $\displaystyle \frac{3}{7} $, $\displaystyle \frac{6}{7} $, $\displaystyle \frac{9}{7} $ > - Feb 15th 2008, 12:58 AMmr fantastic