Equation of a Plane in Vector and Cartesian Form with Formula and Solved Examples
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 in Three Dimensional Geometry
1. What is the equation of a plane in 3D geometry?
The equation of a plane in 3D geometry is most commonly written as Ax + By + Cz + D = 0, where A, B, and C are constants. This form is called the general equation of a plane.
- A, B, and C are the components of the normal vector to the plane.
- D is a constant that shifts the plane in space.
- It represents all points (x, y, z) lying on the plane.
This equation is widely used in coordinate geometry and vector algebra.
2. What is the formula for the equation of a plane passing through a point?
The equation of a plane passing through a point (x₁, y₁, z₁) with normal vector (A, B, C) is A(x − x₁) + B(y − y₁) + C(z − z₁) = 0.
- (x₁, y₁, z₁) is a known point on the plane.
- (A, B, C) is the normal vector.
This is called the point-normal form of the plane equation and is useful when a point and direction are given.
3. How do you find the equation of a plane given three points?
To find the equation of a plane through three non-collinear points, first compute a normal vector using the cross product, then use the point-normal form.
- Step 1: Form two direction vectors from the three points.
- Step 2: Find their cross product to get the normal vector.
- Step 3: Substitute into A(x − x₁) + B(y − y₁) + C(z − z₁) = 0.
The three points must not lie on the same straight line.
4. What is the intercept form of the equation of a plane?
The intercept form of a plane is x/a + y/b + z/c = 1, where a, b, and c are the intercepts on the coordinate axes.
- a is the x-intercept.
- b is the y-intercept.
- c is the z-intercept.
This form is useful when the plane cuts all three axes at known points.
5. What is the normal vector of a plane?
The normal vector of a plane is a vector perpendicular to the plane and is given by (A, B, C) in the equation Ax + By + Cz + D = 0.
- It determines the orientation of the plane.
- Any vector parallel to (A, B, C) is also normal to the plane.
The normal vector is essential for finding angles and distances involving planes.
6. How do you find the distance from a point to a plane?
The distance from a point (x₁, y₁, z₁) to a plane Ax + By + Cz + D = 0 is |Ax₁ + By₁ + Cz₁ + D| / √(A² + B² + C²).
- Substitute the point into the plane equation.
- Take the absolute value.
- Divide by √(A² + B² + C²).
This formula is commonly used in 3D coordinate geometry.
7. How do you check if a point lies on a plane?
A point lies on a plane if it satisfies the equation Ax + By + Cz + D = 0.
- Substitute the coordinates (x, y, z) into the equation.
- If the result equals 0, the point lies on the plane.
- If not, the point is outside the plane.
This method is straightforward and commonly tested in exams.
8. What is the angle between two planes?
The angle between two planes is the angle between their normal vectors and is given by cosθ = (A₁A₂ + B₁B₂ + C₁C₂) / (√(A₁²+B₁²+C₁²) √(A₂²+B₂²+C₂²)).
- (A₁, B₁, C₁) and (A₂, B₂, C₂) are the normal vectors.
- Use the dot product formula.
If the dot product is zero, the planes are perpendicular.
9. How do you find the equation of a plane parallel to another plane?
A plane parallel to Ax + By + Cz + D = 0 has the same normal vector and is written as Ax + By + Cz + D₁ = 0.
- The coefficients A, B, and C remain the same.
- Only the constant term changes.
Parallel planes have identical normal vectors but different positions in space.
10. Can you give an example of finding the equation of a plane?
The equation of a plane through point (1, 2, 3) with normal vector (2, −1, 4) is 2(x − 1) − (y − 2) + 4(z − 3) = 0.
- Use the point-normal form: A(x − x₁) + B(y − y₁) + C(z − z₁) = 0.
- Substitute A=2, B=−1, C=4 and (1,2,3).
- Simplified form: 2x − y + 4z − 12 = 0.
This is a standard worked example in 3D coordinate geometry.
