- #1
forty
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Find the distance of the point P(-1, 2, -1) from the line whose parametric equations are:
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)
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)
Last edited: