What is the Equation of a Plane (Line & Point)? | Class 12

S7-SA2-0281

What is the Equation of a Plane Containing a Line and a Point Not on the Line?

Grade Level:

Class 12

AI/ML, Physics, Biotechnology, FinTech, EVs, Space Technology, Climate Science, Blockchain, Medicine, Engineering, Law, Economics

Definition

What is it?

The equation of a plane containing a line and a point not on that line describes a flat, two-dimensional surface in 3D space. This plane is uniquely determined because the line gives two points (or a point and a direction), and the additional point provides the third non-collinear point needed to define a plane.

Simple Example

Quick Example

Imagine you have a straight, long cricket bat lying on the ground (this is your line). Now, place a single cricket ball somewhere else on the ground, not on the bat (this is your point). The flat surface of the ground itself, which holds both the bat and the ball, represents the plane whose equation we are trying to find.

Worked Example

Step-by-Step

Find the equation of the plane containing the line (x-1)/2 = (y+1)/3 = (z-2)/4 and the point (0, 1, 0).

Step 1: Identify a point on the line and the direction ratios of the line. From (x-1)/2 = (y+1)/3 = (z-2)/4, a point on the line is A(1, -1, 2) and the direction ratios of the line are (2, 3, 4). Let this direction vector be **b** = 2**i** + 3**j** + 4**k**.
---Step 2: Identify the given point not on the line. Let this be P(0, 1, 0).
---Step 3: Find a vector connecting the point on the line A to the given point P. Vector **AP** = P - A = (0-1)**i** + (1-(-1))**j** + (0-2)**k** = -**i** + 2**j** - 2**k**.
---Step 4: The normal vector **n** to the plane is perpendicular to both the direction vector of the line (**b**) and the vector **AP**. So, **n** = **b** x **AP**.
**n** = (2**i** + 3**j** + 4**k**) x (-**i** + 2**j** - 2**k**)
**n** = | i j k |
| 2 3 4 |
|-1 2 -2 |
**n** = **i**((3)(-2) - (4)(2)) - **j**((2)(-2) - (4)(-1)) + **k**((2)(2) - (3)(-1))
**n** = **i**(-6 - 8) - **j**(-4 + 4) + **k**(4 + 3)
**n** = -14**i** + 0**j** + 7**k** = (-14, 0, 7).
---Step 5: Use the normal vector (-14, 0, 7) and either point (A or P) to write the plane equation. Using point P(0, 1, 0), the equation is a(x - x1) + b(y - y1) + c(z - z1) = 0.
-14(x - 0) + 0(y - 1) + 7(z - 0) = 0.
---Step 6: Simplify the equation.
-14x + 7z = 0.
Dividing by 7, we get -2x + z = 0 or 2x - z = 0.
---Answer: The equation of the plane is 2x - z = 0.

Why It Matters

Understanding plane equations is super important in fields like Engineering and AI/ML. For example, in self-driving cars, this helps computers map out roads and obstacles, like finding the 'flat' surface of the road. It's also used in creating realistic 3D graphics for games and movies!

Common Mistakes

MISTAKE: Assuming the given point IS on the line. | CORRECTION: Always verify if the point satisfies the line's equation. If it does, you have two points on the line, but not a unique line and a separate point.

MISTAKE: Incorrectly calculating the cross product for the normal vector. | CORRECTION: Double-check your cross product calculation. Remember **i**, **j**, **k** signs and the determinant formula.

MISTAKE: Using the direction vector of the line as one of the vectors for the plane equation directly. | CORRECTION: You need a normal vector (perpendicular to the plane). This is found by the cross product of two vectors *within* the plane (like the line's direction vector and a vector connecting the given point to a point on the line).

Practice Questions

Try It Yourself

QUESTION: Find the equation of the plane containing the line x/1 = y/2 = z/3 and the point (1, 0, 0). | ANSWER: 3x - 3y + z - 3 = 0

QUESTION: A plane passes through the line (x-2)/1 = (y-3)/-1 = (z-1)/2 and the point (0, 0, 0). What is its equation? | ANSWER: x + 5y + 2z = 0

QUESTION: Determine the equation of the plane that includes the line passing through (1, 2, 3) and (3, 4, 5), and also contains the point (0, 0, 0). | ANSWER: x - 2y + z = 0

MCQ

Quick Quiz

Which of these is a correct first step to find the equation of a plane containing a line and a point not on the line?

Find the distance between the line and the point.

Identify a point on the line and the direction vector of the line.

Calculate the angle between the line and the point.

Check if the line is parallel to the point.

The Correct Answer Is:

To define a plane, you need points and directions. Identifying a point on the line and its direction vector gives you two crucial pieces of information for constructing the plane's equation. The other options are not relevant or are incorrect first steps.

Real World Connection

In the Real World

In building smart cities, urban planners use such concepts to model the 'planes' of roads or building facades in 3D maps. For instance, when planning a new flyover, engineers at NHAI use these equations to make sure the structure (line) fits perfectly within the overall design (plane) relative to existing landmarks (points).

Key Vocabulary

Key Terms

PLANE: A flat, two-dimensional surface in 3D space that extends infinitely | DIRECTION RATIOS: Numbers proportional to the direction cosines of a line, indicating its orientation in space | NORMAL VECTOR: A vector perpendicular to a plane, indicating its orientation | CROSS PRODUCT: A vector operation that results in a vector perpendicular to the two input vectors

What's Next

What to Learn Next

Great job understanding this! Next, you can explore 'Equation of a Plane Passing Through Three Non-Collinear Points'. It builds on this by using three points to define a plane, which is closely related to how a line and a point define it!