Find the distance of the point P(-1, 2, -1) from the line whose parametric equations are:(adsbygoogle = window.adsbygoogle || []).push({});

x = 2+2t

y = 2-t

z = -3-3t

***

So i worked out the vector equation of the line r = ro + vt

r = (2,2,-3) + t(2,-1,-3)

if i take the dot product of the the point P with the t component of r and equate it to 0 will the value of t give me the point on the line that is closest to P?

(-1,2,-1) . (2+2t,2-t,-3-3t) = 0

-2-2t+4-2t+3+3t = 0

t = 5

So if i use that value in r to get the point (12,-3,-18) is that point the closest point to P? in which case calculating the distance is just the distance from this point to P?

I have a feeling this is all so very very wrong >.<

Thanks :)

************EDIT

This is obviously wrong, re-working this out and will repost to see if what i come up with is even logical ;)

*************NEW try

If i put everything relative to the origin ie so the r vector becomes just t(2,-1,-3) and P becomes (-3,0,2). then take some point on the line Q = (x,y,z) that is closest to P, then QP.r should = 0 (perpendicular)

QP = OP - OQ = (-3,0,2) - (x,y,z) = (-3-x,-y,2-z)

then (-3-x,-y,2-z).(2,-1,-3) = 0

which gives -2x + y + 3z = 12

so where to from here? (if my logic is right)

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# Homework Help: Distance of a point from a line (3-D)

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