Question

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

Answer 1

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

Answer 2

please give me the answer im confused

Answer 3

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

Answer 4

means please solv it

Answer 5

can you give me the direction vector from the line?

Answer 6

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

Answer 7

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

Answer 8

(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)

Answer 9

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

Answer 10

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

Answer 11

does this make sense?

Answer 12

thisi is my false attempt

Answer 13

seems its above his level

Answer 14

@amistre64
"seems ur working is above his level

Answer 15

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

Answer 16

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

Answer 17

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

<-14,11,12> and this vector ?

Answer 18

YA

Answer 19

I GOT IT ...YR METHOD

Answer 20

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 :)

Answer 21

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 ;)

Answer 22

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

Answer 23

this is insane

Answer 24

this is another option I have in mind :)