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

[mill]

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.

 

[LCKurtz]

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.

 

[mill]

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.

 

[Mark44]

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.

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.

 

[LCKurtz]

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.

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.

 

[mill]

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

 

[LCKurtz]

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?

 

[mill]

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?

 

[LCKurtz]

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.