How to Find the Equation of a Plane Passing Through Two Parallel Lines

When dealing with the geometry of three-dimensional space, one common task is to determine the equation of a plane that passes through two parallel lines. This article will guide you through the process of finding the equation of the plane, highlighting the mathematical steps involved and providing examples for clarity.

Understanding the Problem

Given two parallel lines, one can define a plane that passes through a point on the first line and contains the direction vectors of both lines. The task is to find the equation of this plane. Let's explore this in more detail with an example.

Example Problem

Consider the two parallel lines:

L_1 : frac{x}{7} frac{y-2}{3} frac{z 1}{5} passing through the point A_1: (10, -2, 1) with directional vector V_1: (7, 3, 5). L_2 : frac{x-1}{7} frac{y-3}{3} frac{z-2}{5} passing through the point A_2: (1, 3, -2) with the same directional vector V_1: (7, 3, 5)

These lines form a plane that passes through A_1 and contains the non-parallel vector A_1A_2: (1, 5, -3).

Step 1: Equation of the Plane Containing the First Line

The equation of a plane containing the first given line frac{x}{7} frac{y-2}{3} frac{z 1}{5} can be written as:

ax - 0 by 2 cz - 1 0…. … 1

Here, we have the condition that:

7a 3b 5c 0 … … 2

Additionally, the known point (1, 3, -2) of the second line must also lie in the plane defined by equation 1. Therefore:

a 5b - 3c 0 … … 3

Solving equations 2 and 3 using the cross multiplication method, we get:

frac{a}{-17} frac{b}{13} frac{c}{16}

Substituting these values of a, b, c in equation 1, we get the required equation of the plane as:

17x - 13y - 16z 10

Step 2: Finding the Normal Vector

Let's get two vectors parallel to the plane:

(1/7, 1/3, 1/5) ((1, 3, -2) - (10, -2, 1)) (-9, 5, -3)

The cross product of these vectors will give us a vector normal to the plane:

text{Normal} begin{pmatrix} 1/7 1/3 1/5 end{pmatrix} times begin{pmatrix} -9 5 -3 end{pmatrix}

Carrying out the cross product, we get:

text{Normal} (-34, 26, 32) sim (-17, 13, 16)

The normal vector is (-17, 13, 16)

The Cartesian equation of the plane through point (1, 3, -2) with normal vector (-17, 13, 16) is:

-17(x - 1) 13(y - 3) 16(z 2) 0

Expanding and simplifying, we get:

-17x 17 13y - 39 16z 32 0

-17x 13y 16z 10 0

Note that the simplified form is:

17x - 13y - 16z 10

Verification

To verify that the point (10, -2, 1) lies on the plane:

17(10) - 13(-2) - 16(1) 170 26 - 16 10

This confirms that the plane equation 17x - 13y - 16z 10 is correct.

Conclusion

By using the cross product to find the normal vector and the point-to-plane equation, we can efficiently determine the equation of the plane passing through two parallel lines. This method is both systematic and reliable, ensuring accuracy in your geometric solutions.