Distance of a point from a line (3-D)

In summary, the distance of the point P(-1, 2, -1) from the line whose parametric equations are x = 2+2t, y = 2-t, and z = -3-3t is found by first determining the plane perpendicular to the line that contains the point (-1, 2, -1). The point of intersection between this plane and the line is then used to calculate the distance between the point P and the line. This distance is approximately root(19)/7.
  • #1
forty
135
0
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)
 
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  • #2
forty said:
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?
No, it won't and you have no reason to thinks so.

(-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 >.<
It is!

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
gives that as what? What is the significance of that equation? It appears to be the equation of a plane but what plane?

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

The line from the point to the line is, as you say, perpendicular to the line but there is a whole plane of of such lines through any point on the line. You need first to find the plane perpendicular to that line, containing (-1, 2, -1).
But that's easy. As you have already determined, the direction vector of the line is <2, -1, -3>. And that vector is normal to any plane that is normal to the line. So the plane you want is the plane having normal vector <2, -1, -3> and containing point (-1, 2, -1). Of course, that is given by 2(x+1)- (y- 2)- 3(z+ 1)= 0 or 2x-y- 3z+ 1= 0.

Now, where does the given line intersect that plane? Do you see that that point is the point where the perpendicular from (-1, 2, -1) intersects the line? And so the distance between those points is the distance from the point to the line?
 
  • #3
i really need pictures >.<

so what i think your saying is:

point where the plane 2x-y- 3z+ 1= 0 and the line given to me r = (2,2,-3) + t(2,-1,-3) intersect is the point that is the closest to point P. so if i find where that line intersects this plane then just calculate the distance from this point to P that's the answer?
 
  • #4
worked it out to be root(19/7) using pythag theorem

thanks for your help !
 
  • #5
forty said:
worked it out to be root(19/7) using pythag theorem

thanks for your help !
?? Not root(19)/7? Be careful!
 
  • #6
i feel bad saying this... but are you sure?

2 of my friends both worked it out to be the root(19/7)

!_!
 
  • #7
how to find 3D point on the 3D line with given distance and from given 3D point

Hi

how to find 3D point(lie on line) on the 3D line with given distance and from given 3D point(this is also lie in line) where the given line equation

Please explain the derivation
 
  • #8


gbalajimecse said:
Hi

how to find 3D point(lie on line) on the 3D line with given distance and from given 3D point(this is also lie in line) where the given line equation

Please explain the derivation
Rather than hijacking a thread that's over a year old, you should start a brand-new thread.
 
  • #9
forty said:
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)



take the ro of the vector equation and the point given and put it into the formula

(a(x1)+b(y1)+c(z1))/sqrt((a^2)+(b^2)+(c^2))
 

What is the formula for finding the distance of a point from a line in 3-D space?

The formula for finding the distance of a point from a line in 3-D space is:
d = |ax0 + by0 + cz0 + d| / √(a2 + b2 + c2)
where (x0, y0, z0) is the coordinates of the point, and ax + by + cz + d = 0 is the equation of the line.

How is the distance of a point from a line different from the distance of a point from a plane?

The distance of a point from a line is the shortest distance between the point and the line, while the distance of a point from a plane is the shortest distance between the point and any point on the plane.

What is the significance of the absolute value in the formula for finding the distance of a point from a line?

The absolute value in the formula ensures that the distance is always positive, as distance is a measurement of magnitude. This also accounts for the possibility of the point lying on the opposite side of the line from the origin point.

Can the formula for finding the distance of a point from a line be extended to higher dimensions?

Yes, the formula can be extended to any number of dimensions by adding more variables and coefficients to the equation. The general formula for finding the distance of a point from a hyperplane in n-dimensional space is:
d = |a1x1 + a2x2 + ... + anxn + d| / √(a12 + a22 + ... + an2)

What are some real-world applications of finding the distance of a point from a line in 3-D space?

The formula for finding the distance of a point from a line in 3-D space has various applications in fields such as physics, engineering, and computer graphics. It can be used to calculate the shortest distance between two parallel lines, determine the position of a satellite relative to the Earth's surface, and create three-dimensional models of objects in computer graphics.

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