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- anonymous

Could anyone help with this? i got as far as sqrt(4) * sin[coz^-1(sqrt(4)/ 2sqrt(22)] but its wrong? Find the distance the point P(1,5,5) is to the line through the two points
Q(1,2,5), and R(4,4,2).

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- anonymous

- katieb

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- amistre64

|dw:1359730906752:dw|

- amistre64

\[cos\alpha=\frac{u.v}{|u|~|v|}\]
\[\alpha=cos^{-1}\frac{u.v}{|u|~|v|}\]
\[PQ~sin~\alpha=PQ~cos^{-1}\frac{u.v}{|u|~|v|}\]

- anonymous

i had this formula: [((sqrt(a-d)^2 +(b-e)^2+(c-f)^2 * sin(c0s^-1([(a-d)g + (b-e)h + (c-f)i] / sqrt(a-d)^2 +(b-e)^2+(c-f)^2 sqrtg^2+h^2=i^2]))

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- amistre64

i never remember formulas to well so I tend to have to reconstruct a method that will work out :) that was the simplest one I could manufacture :)

- anonymous

i dont understand your method as we did the other one! could you show me how to use yours?

- amistre64

I know the basic formula to fine the angle between QR and PQ, i know that the distance is a perp line value from P to R; which is equal to the hypotenuse times the sine of the angle.
P(1,5,5)
-Q(1,2,5)
--------
<0,3,0>; PQ = 3
Q(1,2,5)
-R(4,4,2).
----------
<-3,-2,3>; QR = sqrt(22)
im curious if they gave us a rt triangle to begin with now :)

- amistre64

P(1,5,5)
-R(4,4,2)
---------
<-3,1,3>
<0,3,0>pq <-3,-2,3>qr
<-3,1,3>pr <-3,1,3>pr
----------------------
0+3+0 9+9-2
<0,3,0>pq
<-3,-2,3>qr
------------
0-6+0 nope, nothings at right angles to begin with :)

- amistre64

\[3~cos^{-1}\frac{-6}{3\sqrt{22}}=abt~6.034\] is the distance from the line to the given point.

- amistre64

|dw:1359731838243:dw|

- anonymous

so it should be 3sqrt22 not 2sqrt22?

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