Distance from a point in space to a plane passing through a point

In summary, to find the distance from the point A = (1, 0, 2) to the plane passing through the point (1, 2, 1) and perpendicular to the line given by the parametric equations x = 7, y = 1 + 2t, z = t - 3, you can use the formula d = | PS * n/|n||, where P is a point on the plane, S is the given point A, and n is the normal vector to the plane. The normal vector can be found by using the direction vector of the given line, which is perpendicular to the plane. From there, you can use the dot product to find the distance from
  • #1
mill
72
0

Homework Statement



Find the distance from the point A = (1, 0, 2) to the plane passing through the point (1, 2, 1) and perpendicular to the line given by the parametric equations x = 7, y = 1 + 2t, z = t - 3.

Homework Equations



d = | PS * n/|n||, equation of a plane,

The Attempt at a Solution


I tried to find the plane with the point (1,2,1) and the parametric equations first.

I used the given <x,y,z> as the normal vectors. n=<A,B,C> and Ax+By+Cz=0. The end result has a t variable which I don't know how to get rid of. 7x+y+2ty+tz+z=0

Following this, I would have used the distance from the point to a plane formula but the plane equation is not in a form that I can use.
 
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  • #2
mill said:

Homework Statement



Find the distance from the point A = (1, 0, 2) to the plane passing through the point (1, 2, 1) and perpendicular to the line given by the parametric equations x = 7, y = 1 + 2t, z = t - 3.

Homework Equations



d = | PS * n/|n||, equation of a plane,

The Attempt at a Solution


I tried to find the plane with the point (1,2,1) and the parametric equations first.

I used the given <x,y,z> as the normal vectors. n=<A,B,C> and Ax+By+Cz=0. The end result has a t variable which I don't know how to get rid of. 7x+y+2ty+tz+z=0

Following this, I would have used the distance from the point to a plane formula but the plane equation is not in a form that I can use.

If you have a point ##P## on the plane, a point ##Q## not on the plane, and a normal vector for the plane ##\vec N##, let ##\vec V = \overline{PQ}##. Then the distance from ##Q## to the plane is$$
d = \frac 1 {|N|}|(\vec V \cdot \vec N)|$$You have everything you need given to you in the problem.
 
  • #3
LCKurtz said:
If you have a point ##P## on the plane, a point ##Q## not on the plane, and a normal vector for the plane ##\vec N##, let ##\vec V = \overline{PQ}##. Then the distance from ##Q## to the plane is$$
d = \frac 1 {|N|}|(\vec V \cdot \vec N)|$$You have everything you need given to you in the problem.

Thank you for taking the time to reply, but from what I see, I think that is just the distance formula I already posted. I know how to use it. What I don't understand is how the variable t figures out in the problem.
 
  • #4
mill said:

Homework Statement



Find the distance from the point A = (1, 0, 2) to the plane passing through the point (1, 2, 1) and perpendicular to the line given by the parametric equations x = 7, y = 1 + 2t, z = t - 3.

Homework Equations



d = | PS * n/|n||, equation of a plane,

The Attempt at a Solution


I tried to find the plane with the point (1,2,1) and the parametric equations first.

I used the given <x,y,z> as the normal vectors.
What does this mean? You weren't given any vectors.

A vector with the same direction of the line is <0, 2, 1>, and a point P(7, 1, -3) lies on the line.

One problem with formulas such as d = | PS * n/|n|| (that you show) is that there is the temptation to plug stuff into it without understanding what you're doing.

See if you can use the given information to find the equation of the plane. That would be a good start.
mill said:
n=<A,B,C> and Ax+By+Cz=0. The end result has a t variable which I don't know how to get rid of. 7x+y+2ty+tz+z=0

Following this, I would have used the distance from the point to a plane formula but the plane equation is not in a form that I can use.
 
  • #5
LCKurtz said:
If you have a point ##P## on the plane, a point ##Q## not on the plane, and a normal vector for the plane ##\vec N##, let ##\vec V = \overline{PQ}##. Then the distance from ##Q## to the plane is$$
d = \frac 1 {|N|}|(\vec V \cdot \vec N)|$$You have everything you need given to you in the problem.

mill said:
Thank you for taking the time to reply, but from what I see, I think that is just the distance formula I already posted. I know how to use it. What I don't understand is how the variable t figures out in the problem.

Do you see how to get a normal vector to the plane from what you are given? That's all you need and you will see what ##t## has to do with it.
 
  • #6
LCKurtz said:
Do you see how to get a normal vector to the plane from what you are given? That's all you need and you will see what ##t## has to do with it.

That is what I don't see. I tried to find the plane in order to find the normal vector, but the normal vector has a t that I can't figure out how to get rid of. 7x+y+2ty+tz+z=0
 
  • #7
mill said:
That is what I don't see. I tried to find the plane in order to find the normal vector, but the normal vector has a t that I can't figure out how to get rid of. 7x+y+2ty+tz+z=0

You don't need the equation of the plane at all. You are given a line perpendicular to the plane. Won't a direction vector for that line be perpendicular to the plane?
 
  • #8
LCKurtz said:
You don't need the equation of the plane at all. You are given a line perpendicular to the plane. Won't a direction vector for that line be perpendicular to the plane?

I now see about the n. In that case would the formula for the distance from a point to a line in space be more appropriate?
 
  • #9
mill said:
I now see about the n. In that case would the formula for the distance from a point to a line in space be more appropriate?

No. Use the method in post #5.
 

FAQ: Distance from a point in space to a plane passing through a point

1. What is the formula for finding the distance from a point in space to a plane passing through a point?

The formula for finding the distance from a point in space to a plane passing through a point is d = |ax + by + cz + d|/√(a^2 + b^2 + c^2), where (x, y, z) is the coordinates of the point and a, b, c, and d are the coefficients of the plane's equation.

2. How do you interpret the result of the distance calculation?

The result of the distance calculation represents the perpendicular distance from the given point to the plane. This means that the shortest possible distance between the point and the plane is represented by the calculated value.

3. Can the distance be negative?

No, the distance cannot be negative. The absolute value in the formula ensures that the result is always positive.

4. What does it mean if the distance is 0?

If the distance is 0, it means that the given point lies on the plane. This can also be interpreted as the point being a part of the plane's surface or lying in the same plane as the given point.

5. Is it possible for the distance to be infinite?

Yes, it is possible for the distance to be infinite. This occurs when the given point is parallel to the plane, meaning that there is no perpendicular distance between them. In this case, the plane and the point are said to be "coincident".

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