Equation of Plane Parallel to Two Lines | Simple Explanation for Class 12

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What is the Equation of a Plane Parallel to Two Given Lines?

Grade Level:

Class 12

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Definition

What is it?

The equation of a plane parallel to two given lines is found when the normal vector (a line perpendicular to the plane) is perpendicular to the direction vectors of both given lines. This means the normal vector is the cross product of the direction vectors of the two lines. The plane's equation is then determined using this normal vector and a point it passes through.

Simple Example

Quick Example

Imagine you have two parallel railway tracks (your two given lines). You want to build a flat bridge (your plane) over them, such that the bridge is perfectly level and doesn't tilt towards either track. The direction of the bridge's support beams (normal vector) would be perpendicular to both tracks, ensuring it's parallel to the tracks' 'direction'.

Worked Example

Step-by-Step

Find the equation of a plane parallel to the lines (x-1)/2 = (y+3)/-1 = z/3 and x/1 = (y-2)/2 = (z+1)/-1, and passing through the point (1, 2, 3).

Step 1: Identify the direction vectors of the given lines. For L1: (x-1)/2 = (y+3)/-1 = z/3, the direction vector is v1 = (2, -1, 3). For L2: x/1 = (y-2)/2 = (z+1)/-1, the direction vector is v2 = (1, 2, -1).

---Step 2: The normal vector (n) to the plane is perpendicular to both v1 and v2. So, n is the cross product of v1 and v2.
n = v1 x v2 = (2, -1, 3) x (1, 2, -1)

---Step 3: Calculate the cross product.
n = ((-1)(-1) - (3)(2), (3)(1) - (2)(-1), (2)(2) - (-1)(1))
n = (1 - 6, 3 - (-2), 4 - (-1))
n = (-5, 5, 5)

---Step 4: We can simplify the normal vector by dividing by 5 (or -5) to get n' = (-1, 1, 1). This doesn't change the direction.

---Step 5: The equation of a plane is Ax + By + Cz + D = 0, where (A, B, C) is the normal vector. Using n' = (-1, 1, 1), the equation is -x + y + z + D = 0.

---Step 6: The plane passes through the point (1, 2, 3). Substitute these coordinates into the plane equation to find D.
-(-1)(1) + (1)(2) + (1)(3) + D = 0
-1 + 2 + 3 + D = 0
4 + D = 0
D = -4

---Step 7: Substitute D back into the plane equation.
-x + y + z - 4 = 0

Answer: The equation of the plane is -x + y + z - 4 = 0 (or x - y - z + 4 = 0).

Why It Matters

Understanding planes and their relationships with lines is crucial in many fields. In AI/ML, it helps define decision boundaries for classifying data. Engineers use it to design structures like bridges or aircraft wings, and in robotics, it guides robot movement and path planning. It's like finding the perfect flat surface for a drone to land on!

Common Mistakes

MISTAKE: Directly using the direction vectors as the normal vector components (A, B, C) for the plane equation. | CORRECTION: Remember the normal vector must be PERPENDICULAR to the lines. This is achieved by taking the CROSS PRODUCT of the direction vectors of the lines.

MISTAKE: Forgetting to substitute the given point into the plane equation to find the constant 'D'. | CORRECTION: After finding the normal vector (A, B, C), set up Ax + By + Cz + D = 0. Then, plug in the coordinates (x, y, z) of the point the plane passes through to correctly calculate D.

MISTAKE: Incorrectly calculating the cross product of the two direction vectors. | CORRECTION: Double-check your cross product calculation. A simple way to remember is: i(v1y*v2z - v1z*v2y) - j(v1x*v2z - v1z*v2x) + k(v1x*v2y - v1y*v2x). Pay close attention to signs.

Practice Questions

Try It Yourself

QUESTION: Find the normal vector to a plane that is parallel to the lines with direction vectors (1, 0, -2) and (3, -1, 1). | ANSWER: (-2, -7, -1)

QUESTION: A plane is parallel to the lines x/1 = y/2 = z/3 and (x-1)/-1 = (y+2)/0 = (z-3)/2. If it passes through the origin (0,0,0), what is its equation? | ANSWER: 4x - 5y + 2z = 0

QUESTION: Find the equation of the plane that contains the point (2, -1, 3) and is parallel to the lines given by the equations: L1: r = (i + j) + lambda(2i - j + k) and L2: r = (2i - k) + mu(i + 2j - 3k). | ANSWER: x - 7y - 5z + 6 = 0

MCQ

Quick Quiz

If a plane is parallel to two lines, what is the relationship between the plane's normal vector and the direction vectors of the lines?

The normal vector is parallel to both direction vectors.

The normal vector is the sum of the direction vectors.

The normal vector is perpendicular to both direction vectors.

The normal vector is collinear with one of the direction vectors.

The Correct Answer Is:

For a plane to be parallel to two lines, its normal vector (which is perpendicular to the plane itself) must be perpendicular to the direction of both lines. This is why we use the cross product to find it.

Real World Connection

In the Real World

Imagine a drone delivering a package in a city like Bengaluru. The drone needs to fly along a path (like a line) and then land on a flat rooftop (a plane). If there are two specific wind currents (like two lines) that could affect its flight, the drone's navigation system, using principles like these, calculates a landing path (a plane) that is perfectly aligned and stable relative to those wind directions, ensuring a safe delivery.

Key Vocabulary

Key Terms

PLANE: A flat, two-dimensional surface that extends infinitely in all directions. | NORMAL VECTOR: A vector that is perpendicular (at 90 degrees) to a plane. | DIRECTION VECTOR: A vector that shows the direction of a line. | CROSS PRODUCT: A mathematical operation on two vectors that results in a third vector perpendicular to both original vectors. | PARALLEL: Objects that are side-by-side and have the same distance continuously between them, never intersecting.

What's Next

What to Learn Next

Great job learning about planes parallel to lines! Next, you can explore finding the angle between a line and a plane, or the distance of a point from a plane. These build on understanding normal vectors and will help you solve even more complex 3D geometry problems, just like understanding how to make a perfect chai leads to mastering different tea blends!