# Finding the Equation for a Plane Containing a Point and Line

• happykamper21
In summary, to find the equation for the plane containing the point <3, -2, 4> and the line given by (x-3)/2 = (y+1)/-1, z=5 (in symmetric form), you need to first find two linearly independent vectors within the plane. This can be done by selecting an arbitrary point on the line and using the tangent vector of the line, along with the difference vector between the arbitrary point and the given point <3, -2, 4>. Once you have these two vectors, you can use a cross product to find the normal vector, which can then be used to determine the equation of the plane.
happykamper21

## Homework Statement

Problem: Please find an equation for the plane that contains the point <3, -2, 4> and that includes the line given by (x-3)/2 = (y+1)/-1, z=5 (in symmetric form). Simplify

## Homework Equations

I'm really not sure where to start and what process to take to arrive to my answer.

## The Attempt at a Solution

My first attempt at this problem was to turn the symmetric equation into a vector equation <2t-3, -t-1, and 5> then take the normal vector <2, -1, 0> but then I'm not sure if I should be doing that in the first place.

What you have done is one of the steps in a possible way of finding a solution. However, it is not clear to me that you have a plan of attack. You need to ask yourself what information you need to define a plane and then try to extract that information from the problem. For example, what is your idea behind finding the tangent vector of the line?

Orodruin said:
What you have done is one of the steps in a possible way of finding a solution. However, it is not clear to me that you have a plan of attack. You need to ask yourself what information you need to define a plane and then try to extract that information from the problem. For example, what is your idea behind finding the tangent vector of the line?
So I know that I need 2 vectors for a plane and currently I'm given a point and a equation of the line. This is the point where I'm unsure of what to do since I don't know how to utilize the tangent vector of the line.

So you have one vector in the plane. Can you think of a way to get a second (linearly independent) based on the information you have available? What are you planning to do once you have two vectors in the plane?

Orodruin said:
So you have one vector in the plane. Can you think of a way to get a second (linearly independent) based on the information you have available? What are you planning to do once you have two vectors in the plane?
Could I possibly set t = 1 and get a point for the line and then use the point and the result I got from t=1 to get a vector? I know I want to get the normal vector from two vectors to get the equation of the plane. Not sure how to get the second vector that's linearly independent. (Just to clarify: the first vector on the plane is the line right?)

happykamper21 said:
Could I possibly set t = 1 and get a point for the line and then use the point and the result I got from t=1 to get a vector?
Does that vector necessarily lead to a tangent vector to the plane?

happykamper21 said:
Just to clarify: the first vector on the plane is the line right?
More accurately, its tangent vector. If the line is to be in the plane, its tangent vector must be parallel to the plane.

Orodruin said:
Does that vector necessarily lead to a tangent vector to the plane?More accurately, its tangent vector. If the line is to be in the plane, its tangent vector must be parallel to the plane.

So I've thought about this a lot more. This is my thought process. P1: (3,-2,4), P2: (3, -1, 5), P3: t=1: (5, -2, 5). I can then find the vector P1 to P2 and P1 to P3. Use the two vectors I found to do a cross product which will give me the normal vector, which I can then use to come up with the equation of the plane.

happykamper21 said:
So I've thought about this a lot more. This is my thought process. P1: (3,-2,4), P2: (3, -1, 5), P3: t=1: (5, -2, 5). I can then find the vector P1 to P2 and P1 to P3. Use the two vectors I found to do a cross product which will give me the normal vector, which I can then use to come up with the equation of the plane.
This is a valid procedure (I assume you got P2 and P3 by picking two points on the line). Note that you could just as well replace either of the two vectors you propose by the vector from P2 to P3, which will just be (proportional to) the tangent vector of the line. I would probably just have taken an arbitrary point on the line and used the tangent vector of the line and the difference vector from the arbitrary point to P1.

Orodruin said:
This is a valid procedure (I assume you got P2 and P3 by picking two points on the line). Note that you could just as well replace either of the two vectors you propose by the vector from P2 to P3, which will just be (proportional to) the tangent vector of the line. I would probably just have taken an arbitrary point on the line and used the tangent vector of the line and the difference vector from the arbitrary point to P1.
Thank you so much for your patience and time Orodruin. Been stuck on this problem for a long time.

happykamper21 said:
So I've thought about this a lot more. This is my thought process. P1: (3,-2,4), P2: (3, -1, 5), P3: t=1: (5, -2, 5). I can then find the vector P1 to P2 and P1 to P3. Use the two vectors I found to do a cross product which will give me the normal vector, which I can then use to come up with the equation of the plane.

What is stopping you from doing that?

## 1. What is the equation for a plane in 3-dimensional space?

The equation for a plane in 3-dimensional space is Ax + By + Cz = D, where A, B, and C are the coefficients of the x, y, and z variables, and D is a constant.

## 2. How many variables are there in the equation for a plane?

There are three variables in the equation for a plane: x, y, and z. These represent the coordinates of a point in 3-dimensional space.

## 3. What does the D constant represent in the equation for a plane?

The D constant in the equation for a plane represents the distance of the plane from the origin. It is the perpendicular distance from the origin to the plane.

## 4. Can the equation for a plane be written in different forms?

Yes, the equation for a plane can also be written as A(x-x0) + B(y-y0) + C(z-z0) = 0, where (x0, y0, z0) is a point on the plane. This form is known as the "point-normal form".

## 5. How is the equation for a plane related to linear algebra?

The equation for a plane is a fundamental concept in linear algebra, as it represents a two-dimensional subspace in a three-dimensional space. It is used to solve systems of linear equations and to study the properties of vectors and matrices.

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