Mathematics
OpenStudy (anonymous):

find the distance of the point (-1,3,9) from a line (x-13)/5 = (y+8)/-6 = (z+3)/1 plz reply as soon as possible

OpenStudy (amistre64):

since we have a direction vector; we could simply create another vector from a pointon the line to the given point to do operations on

OpenStudy (anonymous):

OpenStudy (amistre64):

an answer will not subdue confusion .... only exacerbate it

OpenStudy (amistre64):

|dw:1331656348406:dw|

OpenStudy (anonymous):

OpenStudy (amistre64):

can you give me the direction vector from the line?

OpenStudy (anonymous):

leet e tell u how did i do it ok???

OpenStudy (amistre64):

that would be great in the mean time I will prepare my method for comparison

OpenStudy (amistre64):

(x-13)/5 = (y+8)/-6 = (z+3)/1 = t therefore: (x-13)/5 = t (y+8)/-6 = t (z+3) /1 = t x-13 = 5t x = 13+5t y+8 = -6t y = -8-6t z+3 = 1t z = -3+1t ........................ our parametric line equation is: x = 13+5t y = -8 -6t z = -3+1t ^ ^ ^ ^ ^ vector ^ point our direction vector is: <5,-6,1> and our point to form a vector to our given point is then (13,-8,-3)

OpenStudy (amistre64):

given - line = vector from line to point (-1 , 3, 9) -(13,-8,-3) ------------ <-14,11,12> with this information we can determine the angle between the vectors and the sine of that angle will help us with the distance

OpenStudy (amistre64):

v.u ---- = cos(t) |v||u| v.u cos^-1 (----) = t |v||u| v.u pi/2 - cos^-1 (----) = compliment of t |v||u| |u| cos(comp t) = distance from line

OpenStudy (amistre64):

does this make sense?

OpenStudy (anonymous):

thisi is my false attempt

OpenStudy (aravindg):

seems its above his level

OpenStudy (aravindg):

@amistre64 "seems ur working is above his level

OpenStudy (amistre64):

my thought process might be a bit convoluted, bit the level is pretty much the same :)

OpenStudy (amistre64):

if youve followed my reasoning up to the vectors to play with, the rest is just finger work

OpenStudy (amistre64):

<5,-6,1> do you see how we obtain this vector? <-14,11,12> and this vector ?

OpenStudy (anonymous):

YA

OpenStudy (anonymous):

I GOT IT ...YR METHOD

OpenStudy (amistre64):

good, then lets get our parts that we need from them to use :) v = < 5 ,-6 , 1> |v| = sqrt(5^2+6^2+1^2) = sqrt(62) u = <-14,11,12> |u| = sqrt(14^2+11^2+12^2) = sqrt(461) ---------------- v.u= -70-66+12 = -124 with any luck those are good numbers :)

OpenStudy (amistre64):

cos(t) = -124/sqrt(62)(461)) t = arccos(-124/sqrt(62)(461))) sin(t) = sin(arccos(-124/sqrt(62)(461)))) |u| sin(t) = sqrt(461)*sin(arccos(-124/sqrt(62)(461)))) = distance from point to line http://www.wolframalpha.com/input/?i=sqrt%28461%29*sin%28arccos%28-124%2F%28sqrt%2862%29%28461%29%29%29%29%29 21.458 if were lucky ;)

OpenStudy (amistre64):

im sure theres a simpler way, but i can never remember it if there is

OpenStudy (anonymous):

this is insane

OpenStudy (amistre64):

|dw:1331658488112:dw| this is another option I have in mind :)