A plane is any flat and two-dimensional surface that can extend infinitely in terms of distance. It can also be considered as a two-dimensional analogue of a point that has zero dimensions, a line that has one dimension, and a space that has 3 dimensions.
This plane with a three-dimensional space has the following equation
ax + by + cz + d=0, ax+by+cz+d=0,
In this plane, there should be at the minimum one of the numbers a, b,a,b, and cc must have a value which is non-zero. A plane in this coordinate space can be determined by the use of a point along with a vector that is at a 90-degree angle or perpendicular to the plane.
Equation of Plane in 3d
A plane in 3D coordinate space is established by a point and a vector that is at the angle of 90 degrees to the plane. Let P0=(x0,y0,z0) be the point given, and n as the orthogonal vector. Also, consider P=(x,y,z) as any point in the plane, and r and r0 be the position vectors of points P and P0, respectively. Now if we let, n=(a,b,c), then, since we already know that P0P is at 90 degrees to the n in position, we get
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P0P⋅n=(r−r0)⋅n
=(x−x0,y−y0,z−z0)⋅(a,b,c)
= a(x-x0) + b(y-y0) + c(z-z0)
=0
The above equation of the plane can also be written as
ax+by+cz+d = 0,
where d= -(ax0} + by0 + cz0.d=−(ax0+by0+cz0).
This equation, however, does not hold true if one of a, b, c is zero. Wherein, the vector is parallel to either one of the points of coordinate planes. Let’s Say c = 0, in this case, the vector is parallel to the XY plane and the equation of this plane will be as a(x-x0) + b(y-y0) = 0 which is in a straight line in the plane of XY and z is clear. Similar opinions affect if two of a, b, c is zero.
An Alternative way to contemplate the equation of the plane is as a flattened parallelepiped. A flattened parallelepiped is constructing by the use of three vectors a=⟨x1,y1,z1⟩,b=⟨x2,y2,z2⟩,c=⟨x3,y3,z3⟩, has a definite volume of 0. This volume can also be calculated using the scalar triple product.
0=a⋅(b×c), which gives the vector located as normal to the plane.
Let's now assume that the endpoints of (b×c) are ( x, y, z )(x,y,z) and (x_0, y_0, z_0 )(x0,y0,z0) and the constituents of a= (a,b,c) Then by considering the dot product, the equation we get is,
0=a(x−x0)+b(y−y0)+c(z−z0).
Parallel to the Coordinate Planes
The equation of planes which are parallel to each of the xy-, yz-, and xz-planes and passing through a point A=(a,b,c)A=(a,b,c) is considered as follows:
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1) The equivalence of the plane which is parallel to the xy plane is z=c.
2) The equivalence of the plane which is parallel to the yz plane is x=a
3) The equivalence of the plane which is parallel to the xz plane is y=b.
Let’s try a problem for better understanding:
What is the equation of the plane which passes through the point B=(4,1,0)B=(4,1,0) and is parallel to the yz-plane?
Since the xx-coordinate of B is 4, the equation of the plane passing through B parallel to the yz-plane is
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Let us determine the equation of plane that will pass through given points
(-1,0,1) parallel to the xz plane.
Normal Vector and a Point
The equation of a plane is easily established if the normal vector of a plane and any one point passing through the plane is given.
Thus, the equation of a plane through a point A=(x_{1}, y_{1}, z_{1} )A=(x1,y1,z1) whose normal vector is N = (a,b,c) is
a(x-x_{1}) + b(y- y_{1} )+ c(z-z_{1}) = 0 .a(x−x1)+b(y−y1)+c(z−z1)=0.
Check out the Following Examples:
If a plane is passing through the point A=(1,3,2) and has a normal vector N = (3,2,5), then what is the equation of the plane?
The equation of the plane which passes through A=(1,3,2) and has normal vector N = (3,2,5) is
3(x-1) + 2(y-3) + 5(z-2) =0
3x - 3 + 2y - 6 + 5z - 10 = 0
3x + 2y + 5z - 19 =0.
Q. If a plane is passing through the point A=(5,6,2) and has a normal vector n = (-1,3,-7), then what is the equation of the plane?
The equation of the plane which passes through the point A=(5,6,2)A=(5,6,2) and has normal vector n= (-1,3,-7) is
-1(x-5) + 3(y-6) -7(z-2) = 0
-x+5+3y-18-7z+14= 0
-x+3y-7z+1 =0
How to find the equation of a plane in 3d when three points of the plane are given?
When we know three points on a plane, we can find the equation of the plane by solving simultaneous equations.
Let ax+by+cz+d=0 be the equation of a plane on which there are the following three points: A=(1,0,2), B=(2,1,1), and C=(-1,2,1). Then the equation of the plane is established as follows:
We already have the equation of the plane with 4 unknown constants:
ax + by + cz +d = 0………… (1)
We also get the following 3 equations by substituting the coordinates of A, B, and C into (1):
a⋅1+b⋅0+c⋅2+d=0
a⋅2+b⋅1+c⋅1+d=0
a⋅(−1)+b⋅2+c⋅1+d=0
Substituting (2)into (1) we have
ax + 3ay + 4az -9a= 0
x + 3y + 4z - 9 =0.
Hence, the equation of the plane passing through the three points
A=(1,0,2), B=(2,1,1), and C=(-1,2,1) is
x + 3y + 4z - 9 =0 .x+3y+4z−9=0.
Let’s try on some problem questions to understand the concept better.
A plane is passing through, say, 3 points. The three points are as follows:
points A= (0,0,2),
point B= (1,0,1), and
Point C=(3,1,1),
Keeping these points in the equation what equation of plane can be formed?
Consider an equation of plane as ax+by+cz+d=0……..Take the following equation as 1
Now, since the plane is dealing with 3 points, The equation can further be progressed as
a⋅0+b⋅0+c⋅2+d=0
a⋅1+b⋅0+c⋅1+d=0
a⋅3+b⋅1+c⋅1+d=0
This leaves us with =-2a, c=a, d=-2a. ......Take the following equation as 2
The next step is to substitute equation (2) into equation(1), which give us,
ax + -2ay + az -2a = 0
x -2y + z - 2 =0.
Hence, the equation of the plane passing through the three points A= (0,0,2), B=(1,0,1) and C=(3,1,1) is
x -2y + z - 2 =0.
FAQs on Equation of a Plane
1. What is the general equation of a plane in three-dimensional geometry?
The general equation of a plane in three-dimensional space is Ax + By + Cz + D = 0, where A, B, and C are the direction ratios of the normal to the plane, D is a constant, and (x, y, z) are the coordinates of any point on the plane.
2. How do you find the equation of a plane passing through three given non-collinear points?
To determine the equation of a plane through three non-collinear points (x1, y1, z1), (x2, y2, z2), and (x3, y3, z3):
- Form two vectors using the points.
- Find their cross product to get the normal vector (A, B, C).
- Use the point-normal form: A(x - x1) + B(y - y1) + C(z - z1) = 0.
3. What is the significance of the normal vector in the equation of a plane?
The normal vector of a plane, given by (A, B, C) in the equation Ax + By + Cz + D = 0, determines the orientation of the plane in space. All points on the plane satisfy the condition that the vector from a fixed point to any point on the plane is perpendicular to the normal vector.
4. Can a plane be uniquely determined by a point and a normal vector? How?
Yes, a plane can be uniquely determined by a point (x0, y0, z0) and a normal vector (A, B, C). The equation is: A(x - x0) + B(y - y0) + C(z - z0) = 0. This ensures that every point (x, y, z) on the plane forms a vector with the point that is perpendicular to the normal.
5. What are some important applications of the equation of a plane in real life and competitive exams?
The equation of a plane is widely used in physics (motion analysis), engineering (designing flat surfaces), computer graphics, and architecture. In competitive exams, it is commonly tested for its concepts of distance, intersection, angles, and relations with other planes or lines.
6. How do different forms of the plane's equation—vector, cartesian, and normal—differ in their usage?
The cartesian form (Ax + By + Cz + D = 0) directly relates to coordinate axes and is most commonly used for calculations. The vector form utilizes position vectors and describes the plane using points and direction vectors. The normal form expresses the plane's orientation and distance from the origin, useful for distance calculations and geometric reasoning.
7. What is the condition for two planes to be parallel or perpendicular in space?
- Two planes are parallel if their normal vectors are proportional, that is, A1/A2 = B1/B2 = C1/C2.
- They are perpendicular if the dot product of their normal vectors is zero: A1A2 + B1B2 + C1C2 = 0.
8. What conceptual errors do students most often make while finding the equation of a plane through three points?
Common mistakes include miscalculating vector directions, incorrectly finding the cross product, assuming points are always non-collinear, and omitting to substitute points correctly into the equation. Ensuring vectors are correctly formed and the normal vector is used properly helps avoid these errors.
9. How does the distance of a point from a plane help in solving exam-based geometry problems?
The distance of a point (x1, y1, z1) from a plane Ax + By + Cz + D = 0 is given by the formula:
|A x1 + B y1 + C z1 + D| / √(A² + B² + C²). This is frequently asked in board exams for calculating shortest paths or verifying a point's location relative to a plane.
10. In what ways can multiple planes intersect, and how is the line or point of intersection determined?
- Two planes can intersect in a straight line, found by solving their equations simultaneously.
- Three planes may intersect in a point, a line, or may have no common intersection. The solution set to their equations (if unique) gives the intersection point, otherwise the line or common region.
