closest point

**chaddy** · February 12th 2008, 02:25 PM

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

**ecMathGeek** · February 12th 2008, 02:32 PM

Originally Posted by chaddy

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

To do this, find the normal vector to the plane that passes through the origin.

**chaddy** · February 12th 2008, 02:34 PM

How would I find the normal vector?

**chaddy** · February 14th 2008, 07:34 PM

Does anyone know?

**mr fantastic** · February 14th 2008, 08:00 PM

Originally Posted by chaddy

How would I find the normal vector?

For the plane ax + by + cz = d, a normal vector is ai + bj + ck.

**mr fantastic** · February 14th 2008, 08:06 PM

Originally Posted by mr fantastic

For the plane ax + by + cz = d, a normal vector is ai + bj + ck.

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:

$a(at) + b(bt) + c(ct) = d \Rightarrow t = \frac{d}{a^2 + b^2 + c^2}$ .

Therefore the coordinates of the point in the plane closest to the origin are:

$x = at = \frac{ad}{a^2 + b^2 + c^2}$ .

$y = bt = \frac{bd}{a^2 + b^2 + c^2}$ .

$x = at = \frac{cd}{a^2 + b^2 + c^2}$ .

And as a bonus, the answer to what the closest distance is equal to is ..........

**chaddy** · February 15th 2008, 12:10 AM

$ x = \frac{6}{14} $

$ y = \frac{12}{ 14} $

$ z = \frac{18}{ 14} $

So the point closest to the origin would be < $\frac{3}{7}$ , $\frac{6}{7}$ , $\frac{9}{7}$ >

**mr fantastic** · February 15th 2008, 12:58 AM

Originally Posted by chaddy

$ x = \frac{6}{14} $

$ y = \frac{12}{ 14} $

$ z = \frac{18}{ 14} $

So the point closest to the origin would be < $\frac{3}{7}$ , $\frac{6}{7}$ , $\frac{9}{7}$ >

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