# General question about vectors. (point to plane)

• Apr 26th 2009, 01:11 AM
AshleyT
General question about vectors. (point to plane)
Ok, this is more a general question...There's a picture if that helps for me to explain my question :).

Quote:

1/ Find the perpendicular distance of the point P(4, 5, 6) and the
plane x + 2y - 2z = 0.
This is the example in my text book. So i was reading through how the derive the formula

$\displaystyle PM = \frac{n . p}{| n | }$
Now what i don't understand with this...n is the same vector as PM...
So the modulus of n should be the distance of PM. But then if thats the case, the formula doesn't make sense because it would be saying
$\displaystyle |x| = \frac{x.y}{|x|}$
and we need to know |x| to work out |x|...therefore the formula is pointless...
I know this mustn't be the case and its just me not understanding.

Thanks
• Apr 26th 2009, 05:09 AM
Plato
I think that you have not really understood the proof.
The usual formula for that distance is, using your diagram, $\displaystyle \frac{{\left| {\overrightarrow {PT} \cdot N} \right|}}{{\left\| N \right\|}}$.
But in that the $\displaystyle N$ is the normal to the plane.
That formula is derived by using the cosine of the angle between $\displaystyle \overrightarrow {PT}~\&~N$.
The distance is the length of one side of a right triangle with hypotenuse $\displaystyle \overrightarrow {PT}$.
• Apr 28th 2009, 10:04 AM
AshleyT
Quote:

Originally Posted by Plato
I think that you have not really understood the proof.
The usual formula for that distance is, using your diagram, $\displaystyle \frac{{\left| {\overrightarrow {PT} \cdot N} \right|}}{{\left\| N \right\|}}$.
But in that the $\displaystyle N$ is the normal to the plane.
That formula is derived by using the cosine of the angle between $\displaystyle \overrightarrow {PT}~\&~N$.
The distance is the length of one side of a right triangle with hypotenuse $\displaystyle \overrightarrow {PT}$.

Hi, thanks for the reply. Much appreciated :)