Hey I just have a little concern about the distance between a plane & a point not on the plane.(adsbygoogle = window.adsbygoogle || []).push({});

In the picture here:

[PLAIN]http://img838.imageshack.us/img838/4197/normalyy.jpg [Broken]

we can define the equation of a line passing throughBdown to the plane and beyond as

X=B+ tN

If we want to find the equation of the line passing through the pointPin the plane

through toAwe'll use the idea of the dot product due to it's orthogonal characteristic.

(X-A)•N=0

I want to point out thatXis going to have to be equal to the pointPin the picture so

X=P=B+ tN

and

(X-A)•N= (P-A)•N= (B+ tN-A)•N=0

We'll fineagle it to get

(B+ tN-A)•N=0

(B-A)•N+ tN•N=0

tN•N= - (B-A)•N

tN•N= (A-B)•N

[tex] t \ = \ \frac{ ( \overline{A} \ - \ \overline{B}) \cdot \overline{N}}{ \overline{N} \cdot \overline{N}} [/tex]

Then you can sub this into your original equation

X=B+ tN

To find the actual value at the pointP.

Using this couldn't you just find the norm of this numerical value to find the length of

the line connecting B to P?

Then you could go on to use the General Pythagoras Theorem and get

[tex] || \overline{A} \ + \ \overline{B} || \ = \ || \overline{AP} || \ + \ || \overline{PB} || [/tex]

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# Homework Help: Point-Plane Distance?

Can you offer guidance or do you also need help?

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