How To Find The Equation Of A Plane Passing Through 3 Non Collinear Points With Formula And Examples
A plane is a smooth, two-dimensional surface, which stretches infinitely far. A plane is a two - dimensional representation of a point (zero dimensions), a line (one dimension), and a three-dimensional object. A plane in 3-dimensional space has the equation axe + by + cz + d = 0, where at least one of the coefficients a, b or c must be non-zero.
A vector is a physical quantity for which both direction and magnitude are defined. A position vector basically defines the position of a particular point in a three-dimensional cartesian plane system, with respect to an origin point. Consider a line on a plane. This line has a length and an arrow. Here, the length is the magnitude and the arrowhead shows the direction. Hence, in a plane, a line is a vector.
Perpendicular Planes to Vectors and Points
For one particular point on the vector, however, there is only one unique plane that passes through it and is also perpendicular to the vector. A vector can be thought of as a collection of points. So, for a particular vector, there are infinite planes that are perpendicular to it.
The vector equation for the following image is written as: (\[\overrightarrow{r}\] — \[\overrightarrow{r}_{0}\]). \[\overrightarrow{N}\] = 0, where \[\overrightarrow{r}\] and \[\overrightarrow{r}_{0}\] represent the position vectors. For this plane, the cartesian equation is written as:
\[A (x - x_{1}) + B (y - y_{1}) + C (z - z_{1}) = 0\], where A, B, and C are the direction ratios.
Equation of Plane Passing through 3 Non-Collinear Points
\[P(x_{1},y_{1},z_{1}),~Q(x_{2},y_{2},z_{2}),~and~R(x_{3},y_{3},z_{3})\] are three non-collinear points on a plane.
We know that: ax + by + cz + d = 0 —————(i)
By plugging in the values of the points P, Q, and R into equation (i), we get the following:
\[a(x_{1}) + b(y_{1}) + c(z_{1}) + d = 0\]
\[a(x_{2}) + b(y_{2}) + c(z_{2}) + d = 0\]
\[a(x_{3}) + b(y_{3}) + c(z_{3}) + d = 0\]
Suppose, P = (1,0,2), Q = (2,1,1), and R = (−1,2,1)
Then, by substituting the values in the above equations, we get the following:
a(1) + b(0) + c(2) + d = 0
a(2) + b(1) + c(1) + d = 0
a(-1) + b(2) + c(1) + d = 0
Solving these equations gives us b = 3a, c = 4a, and d = (-9)a. ———————(ii)
By plugging in the values from (ii) into (i), we end up with the following:
ax + by + cz + d = 0
ax + 3ay + 4az−9a
x + 3y + 4z−9
Therefore, the equation of the plane with the three non-collinear points P, Q, and R is x + 3y + 4z−9.
Solved Examples
Example 1: A (3,1,2), B (6,1,2), and C (0,2,0) are three non-collinear points on a plane. Find the equation of the plane.
Solution:
We know that: ax + by + cz + d = 0 —————(i)
By plugging in the values of the points A, B, and C into equation (i), we get the following:
a(3) + b(1) + c(2) + d = 0
a(6) + b(1) + c(2) + d = 0
a(0) + b(2) + c(0) + d = 0
Solving these equations gives us a = 0,
c =12
b, d = —2b ———————(ii)
By plugging in the values from (ii) into (i), we end up with the following:
ax + by + cz + d = 0
0x + (—by) +12bz — 2b = 0
x - y +12z —2 = 0
2x-2y + z-4 = 0
Therefore, the equation of the plane with the three non-collinear points A, B, and C is
2x-2y + z-4 = 0.
Example 2: S (0,0,2), U (1, 0, 1), and V (3, 1,1) are three non-collinear points on a plane. Find the equation of the plane.
Solution:
We know that: ax + by + cz + d = 0 —————(i)
By plugging in the values of the points S, U, and V into equation (i), we get the following:
a(0) + b(0) + c(2) + d = 0
a(1) + b(0) + c(1) + d = 0
a(3) + b(1) + c(1) + d = 0
Solving these equations gives us b = —2a, c = a, d = —2a ———————(ii)
By plugging in the values from (ii) into (i), we end up with the following:
ax + by + cz + d = 0
ax + —2ay + az — 2a = 0
x-2y + z-2 = 0
Therefore, the equation of the plane with the three non-collinear points A, B, and C is
x-2y + z-2 = 0.
FAQs on Equation Of A Plane Passing Through 3 Non Collinear Points Explained
1. What is the equation of a plane passing through three non-collinear points?
The equation of a plane passing through three non-collinear points is found using the formula A(x − x₁) + B(y − y₁) + C(z − z₁) = 0, where (A, B, C) is a normal vector obtained from the cross product of two direction vectors in the plane. If the points are P(x₁, y₁, z₁), Q(x₂, y₂, z₂), and R(x₃, y₃, z₃), then:
- Find direction vectors: PQ and PR
- Compute the cross product PQ × PR to get the normal vector
- Substitute the normal vector and one point into the plane equation
2. How do you find the equation of a plane through three given points step by step?
To find the equation of a plane through three non-collinear points, compute a normal vector using the cross product and substitute into the plane formula. Follow these steps:
- Step 1: Form vectors PQ and PR
- Step 2: Find PQ × PR to get normal vector (A, B, C)
- Step 3: Use formula A(x − x₁) + B(y − y₁) + C(z − z₁) = 0
- Step 4: Simplify to standard form Ax + By + Cz + D = 0
3. Why must the three points be non-collinear to form a plane?
Three points must be non-collinear because collinear points lie on a single line and do not uniquely determine a plane. If the points are collinear:
- The direction vectors are parallel
- The cross product becomes zero
- No unique normal vector exists
4. What is the determinant form of the equation of a plane through three points?
The determinant form of the equation of a plane through three non-collinear points is given by the condition that a 3×3 determinant equals zero. It is written as:
| x − x₁ y − y₁ z − z₁ |
| x₂ − x₁ y₂ − y₁ z₂ − z₁ | = 0
| x₃ − x₁ y₃ − y₁ z₃ − z₁ |
This represents that the vectors are coplanar and provides an alternative way to derive the Cartesian equation of a plane.
5. Can you give an example of a plane passing through three points?
Yes, the plane through points (1,0,0), (0,1,0), and (0,0,1) has equation x + y + z = 1. Steps:
- Form vectors: PQ = (−1,1,0), PR = (−1,0,1)
- Find cross product → normal vector (1,1,1)
- Use plane formula: 1(x−1) + 1(y−0) + 1(z−0) = 0
- Simplify → x + y + z = 1
6. What is the standard form of the equation of a plane?
The standard form of the equation of a plane is Ax + By + Cz + D = 0, where (A, B, C) is the normal vector. In this form:
- A, B, C are constants
- D is a constant term
- (A, B, C) ⟂ plane
7. How do you find the normal vector of a plane through three points?
The normal vector of a plane through three points is found by computing the cross product of two direction vectors in the plane. Steps:
- Form vectors PQ and PR
- Compute PQ × PR
- The result (A, B, C) is the normal vector
8. How can you check if three points are collinear in 3D?
Three points in 3D are collinear if the cross product of their direction vectors is the zero vector. Procedure:
- Form vectors PQ and PR
- Compute PQ × PR
- If result = (0,0,0), points are collinear
9. What is the vector equation of a plane passing through three points?
The vector equation of a plane through three non-collinear points is r = a + s b + t c, where b and c are direction vectors in the plane. If a is the position vector of point P, then:
- b = PQ
- c = PR
- s, t are parameters
10. What are common mistakes when finding the equation of a plane through three points?
Common mistakes when finding the equation of a plane include calculation errors in the cross product and not checking collinearity. Key errors to avoid:
- Incorrect computation of PQ × PR
- Using collinear points
- Substitution errors in A(x − x₁) + B(y − y₁) + C(z − z₁) = 0
- Not simplifying to standard form
