# Equation of plane given two parallel lines?

• Isaac Wiebe
In summary: Step 3: Convert PQ to parametric form. PQ = (3.5, -2, 1)Since these lines are indeed parallel, we have to find another line or vector on the plane. Consider vector PQ, where point P (3, 0, 0) is on L1 and point Q (0.5, -2, 1) is on L2. Consequently, PQ is <-3.5, -2, 1>. The normal vector is found by taking the cross product of PQ and L1, and the answer is 24x - 13y + 34z = 72.
Isaac Wiebe

## Homework Statement

Find the equation that contains the lines:
2x + 3y + 4z = 6, x - 2y + z = 3

and

(2x - 1)/22 = (y + 2)/2 = (1 - z)/3

## Homework Equations

A plane (in point normal form) is defined by a point on the plane and a vector normal to it.
In general, the form of A(x - x0) + B(y - y0) + C(z - z0) = 0.

You can use the cross product to determine a normal vector given multiple lines on the plane.

## The Attempt at a Solution

Step 1: Convert L1 to parametric form. Take the cross product of <2, 3, 4> and <1, -2, 1>
and that is the vector <11, 2, -7>. A point that is on both planes forming the line of intersection is (3, 0, 0). Thus L1: x = 3 + 11T, y = 2T, and z = -7T.

Step 2: Convert L2 to parametric form. x = 11T + 0.5, y = 2T -2 and z = 1 - 7T.

Since these lines are indeed parallel, we have to find another line or vector on the plane. Consider vector PQ, where point P (3, 0, 0) is on L1 and point Q (0.5, -2, 1) is on L2. Consequently, PQ is <-3.5, -2, 1>. The normal vector is found by taking the cross product of PQ and L1, and I am having a difficult time obtaining the correct answer, 24x - 13y + 34z = 72.

Isaac Wiebe said:

## Homework Statement

Find the equation that contains the lines:
2x + 3y + 4z = 6, x - 2y + z = 3

and

(2x - 1)/22 = (y + 2)/2 = (1 - z)/3

## Homework Equations

A plane (in point normal form) is defined by a point on the plane and a vector normal to it.
In general, the form of A(x - x0) + B(y - y0) + C(z - z0) = 0.

You can use the cross product to determine a normal vector given multiple lines on the plane.

## The Attempt at a Solution

Step 1: Convert L1 to parametric form. Take the cross product of <2, 3, 4> and <1, -2, 1>
and that is the vector <11, 2, -7>. A point that is on both planes forming the line of intersection is (3, 0, 0). Thus L1: x = 3 + 11T, y = 2T, and z = -7T.

Step 2: Convert L2 to parametric form. x = 11T + 0.5, y = 2T -2 and z = 1 - 7T.

I don't get that z.

Oops, the original problem read (1 - z) / 7, NOT (1 - z) / 3. My mistake!

Now check your calculations for PQ.

## 1. What is the equation of a plane given two parallel lines?

The equation of a plane given two parallel lines can be found by taking the cross product of the direction vectors of the two lines and using one of the lines as a point on the plane.

## 2. How do you find the direction vectors of two parallel lines?

To find the direction vectors of two parallel lines, take any two points on each line and subtract their coordinates to get the vector. The two direction vectors will be parallel to each other.

## 3. Can the equation of a plane be determined with two non-parallel lines?

No, the equation of a plane can only be determined with two parallel lines. If the lines are not parallel, they will intersect at a point and the plane would not be uniquely defined.

## 4. How many variables are in the equation of a plane given two parallel lines?

There are three variables in the equation of a plane given two parallel lines: x, y, and z. This is because the equation of a plane in three-dimensional space is in the form of ax + by + cz = d, where a, b, and c are the coefficients of the variables and d is a constant.

## 5. Can the equation of a plane be used to find the distance between two parallel lines?

No, the equation of a plane cannot be used to find the distance between two parallel lines. The distance between two parallel lines is equal to the distance between any two points on the lines, which can be found using the distance formula.

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