Find Equation of Plane: [1,2,-1] Intersecting [x-y+2y+5,2x+3y-z-1]

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In summary, the equation of the plane containing the line of intersection between x - y + 2y + 5 and 2x + 3y - z - 1 and parallel to the line segment with normal vector [1, 2, -1] can be found by setting the coefficients of the plane to be multiples of the normal vector. To find the value of k, one can choose any two variables and solve for the third using the equation of the plane. This will result in the same value of k, regardless of the chosen variables.
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emma3001
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I have to find the equation of the plane that contains the line of intersection of x - y + 2y + 5 and 2x + 3y - z - 1 and is parallel to the line segment with normal vector [1, 2, -1].

I thought that I could go find the dot product of the normal vector I have [1, 2, -1] and the A B and C values for one of the two above equations.

x - y + 2y + 5 + k(2x + 3y - z - 1)=0
I rearrange it to get x(1 + 2k) + y(-1 + 3k) + z(2-k) + (5- k)=0. Now how can I find the value of k?
 
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emma3001 said:
I have to find the equation of the plane that contains the line of intersection of x - y + 2y + 5 and 2x + 3y - z - 1 and is parallel to the line segment with normal vector [1, 2, -1].

I thought that I could go find the dot product of the normal vector I have [1, 2, -1] and the A B and C values for one of the two above equations.

x - y + 2y + 5 + k(2x + 3y - z - 1)=0
I rearrange it to get x(1 + 2k) + y(-1 + 3k) + z(2-k) + (5- k)=0. Now how can I find the value of k?

I'm not sure yet about this argument, since I'm a bit unclear about what the problem intends by saying the plane is supposed to be "parallel to a line segment with normal vector <1, 2, -1>. It seems to me they could just say the plane you want has that as a normal vector. This would tell us that the coefficients for the plane you seek should be multiples of 1, 2, and -1.

As for your question, just pick some point that satisfies the equation for the plane

x - y + 2z + 5 = 0 .

Take any values you like for x and y, say, and solve that equation for z. Bingo! -- you have a point in that plane. Pick values for any two of the variables (even 0 and 0) and solve for the remaining variable -- it's a point in the plane. Putting those coordinates into your equation leaves you with one equation in one unknown. No matter what you use, you'll still get the same value for k.
 

1. What is a plane equation?

A plane equation is a mathematical representation of a flat surface in three-dimensional space. It is typically written in the form Ax + By + Cz + D = 0, where A, B, and C are the coefficients of the variables x, y, and z, and D is a constant.

2. How do you find the equation of a plane?

The equation of a plane can be found using three points on the plane or a point and a normal vector. In this case, the given information [1,2,-1] and [x-y+2y+5,2x+3y-z-1] provides two points on the plane. Using these points, we can calculate the normal vector, which is . Then, we can plug in one of the points and the normal vector into the general equation of a plane to get the specific equation of this plane: .

3. How do you determine if a point lies on a plane?

A point lies on a plane if its coordinates satisfy the equation of the plane. In this case, we can plug in the given point [1,2,-1] into the equation to see if it equals 0. If it does, then the point lies on the plane.

4. What is the significance of the normal vector in a plane equation?

The normal vector is perpendicular to the plane and determines the orientation of the plane. It can also be used to find the distance between the plane and a point, as well as the angle between two intersecting planes.

5. Can you graph a plane using its equation?

Yes, a plane can be graphed using its equation. In this case, the equation can be rewritten as . This is the equation of a plane in slope-intercept form, which can be graphed on a 3D coordinate system.

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