1. ## closest point

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

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.

3. How would I find the normal vector?

4. Does anyone know?

How would I find the normal vector?
For the plane ax + by + cz = d, a normal vector is ai + bj + ck.

6. 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 ..........

7. $
x = \frac{6}{14}
$

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

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

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

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

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

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

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