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mags093
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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\[cos\alpha=\frac{u.v}{|u|~|v|}\] \[\alpha=cos^{-1}\frac{u.v}{|u|~|v|}\] \[PQ~sin~\alpha=PQ~cos^{-1}\frac{u.v}{|u|~|v|}\]
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]))
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 :)
i dont understand your method as we did the other one! could you show me how to use yours?
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 :)
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 :)
\[3~cos^{-1}\frac{-6}{3\sqrt{22}}=abt~6.034\] is the distance from the line to the given point.
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so it should be 3sqrt22 not 2sqrt22?