# closest point

• Feb 12th 2008, 03:25 PM
closest point
I need to find the point in the plane x + 2y + 3z = 6 that is nearest the origin.
• Feb 12th 2008, 03:32 PM
ecMathGeek
Quote:

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.
• Feb 12th 2008, 03:34 PM
How would I find the normal vector?
• Feb 14th 2008, 08:34 PM
Does anyone know?
• Feb 14th 2008, 09:00 PM
mr fantastic
Quote:

How would I find the normal vector?

For the plane ax + by + cz = d, a normal vector is ai + bj + ck.
• Feb 14th 2008, 09:06 PM
mr fantastic
Quote:

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 ..........
• Feb 15th 2008, 01:10 AM
$
x = \frac{6}{14}
$

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

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

So the point closest to the origin would be < $\frac{3}{7}$, $\frac{6}{7}$, $\frac{9}{7}$ >
• Feb 15th 2008, 01:58 AM
mr fantastic
Quote:

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

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

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

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

(Yes)