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# help on method for distance from line to point (FP4) watch

1. ok i made this up so the numbers probebly wont come out nicely

point A is (1,7,19)

line L is r=(5,16,-12) + t(9,7,-4)

(the (9,7,-4) is direction vector but i cant use latex thing sorry)

find distance from the point to the line...

please just run through the possible methods 1 so i can see if i did i right, and 2 to see which of the 2 i used i got wrong

will tell u how i did it if u want but i probebly got it wrong both ways lol

thanks ...
2. anyone?
4. method 1

from equation of line, with parametre t=0
=> point B is (5,16,-12)

AB = B-A = (4,9,-31)

i used the scalar product of

AB dot (9 7 -4) all over mod (9 7 -4)

giving 223 over root 146

approx 18.46
5. method 2

point P is a general point on the line where a perpendicular to the line which also passes through the point A intesects the line

AP = OP - OA = ( 9t+4 7t+9 -4t-31 )

BP = t( 9 7 -4 )

then if AP dot BP = 0 ... cos (angle between) = 0 so angle is 90 deg

=> ( 9t+4 7t+9 -4t-31 ) dot ( 9 7 -4 ) = 0

=> 146t + 223 = 0

=> t -223/146

=> mod AP = ( (9(-223/146)+4) (7(-223/146)+9) (-4(-223/146)-31) )

=> mod AP = 1/146(root 1423^2 + 247^2 + 3634^2 )

=> mod AP is approx 26.8

thanks...
6. both need the sketch to understand properly but i cant upload one so any help appretiated im sure im just eplaining my problem poorly though lol
7. A correct method (I haven't looked at either of yours) would be to use the point 5,16,-12 on the line (call it B) and find the distance from it to A. This is AB.

Now find the vector to A from B and the vector from B to another point on the line. Use the scalar product to find the angle.

Now use sin(angle) = O/AB where O is the shortest distance.
8. definatly the second method. i got 26.8 when i did it my way

yo do (modPAxPB)/modPB

i j k
9 7 -4 = -181i + 263j + 53k = AP
4 9 -31

(32761+69169+2809)=(104739)

root of (104739) divided by root of (146) = 26.8

BAM lol

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