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
|dw:1331656348406:dw|
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 ?
YA
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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