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
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
please give me the answer im confused
an answer will not subdue confusion .... only exacerbate it
means please solv it
can you give me the direction vector from the line?
leet e tell u how did i do it ok???
that would be great in the mean time I will prepare my method for comparison
(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)
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
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
does this make sense?
thisi is my false attempt
seems its above his level
@amistre64 "seems ur working is above his level
my thought process might be a bit convoluted, bit the level is pretty much the same :)
if youve followed my reasoning up to the vectors to play with, the rest is just finger work
<5,-6,1> do you see how we obtain this vector? <-14,11,12> and this vector ?
I GOT IT ...YR METHOD
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 :)
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 ;)
im sure theres a simpler way, but i can never remember it if there is
this is insane
|dw:1331658488112:dw| this is another option I have in mind :)
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