[Maths Class Notes] on Equation Plane Pdf for Exam

Part of three-dimensional coordinate geometry, a plane is a flat, 2D surface that stretches put infinitely far. A plane of the equation is a 2D analogue of a point having (zero dimensions), a line (one dimension), and three-dimensional space. A plane in 3 D space contains the following equation.

ax + by + cz + d=0,

in which a minimum of one of the numbers a, b, and c should be non-zero. A plane in a 3-dimensional coordinate space is identified by a point and a vector that is perpendicular to the plane.

Equation of a Plane In 3D

An equation plane in 3D coordinate space is identified by a point and a vector which is perpendicular to the plane. Moreover, suppose that P =(x,y,z​) be any point in the plane, and r and r₀ be the position vectors of points P and P0​, respectively. Now, if we let [overrightarrow{n}] = a,b,c , then since [overrightarrow{P0P}] is perpendicular to [overrightarrow {n}],we have;

[overrightarrow {P0​P}]​⋅n​= (r−r₀) ⋅n

=(x−x₀​,y−y₀​,z−z₀​)⋅(a,b,c)

=a(x−x₀​)+b(y−y₀​)+c(z−z₀​)

=0.​

We can also express the above equation of the plane as:

Ax + by + cz + d = 0

In which d= – (ax₀ + by₀ + cz₀)

This does not quite work if one of a, b, c is equivalent to zero. In such a case, that vector is parallel to one of the coordinate planes. Let’s say c = 0, then the vector is parallel to the xy-plane and the equation of the needed plane is a(x-x₀) + b(y-y₀) = 0 which is of course a straight line in the xy plane and z is unrestricted. Likewise, arguments apply if two of a, b, c are zero.

Another way to imagine the vector equation of a plane in 3d is as a flattened parallelepiped. A flattened parallelepiped, composed of three vectors and has a volume equals to 0 Thus, refer to the image shown for the equation of scalar triple product we can use in order to calculate this volume:

Now, Suppose that the endpoints of a vector are ( x, y, z )(x,y,z) and (x₀​,y₀​,z₀​) and the components of vector are ⟨a,b,c⟩. Then by taking the dot product, we obtain the equation of a plane, which is as below:

0 = a(x-x₀) + b(y-y₀) + c (z-z₀)

Solved Examples

Example:

If a plane is crossing across the point A= (1, 3, 2) and has a normal vector [overrightarrow{n}] = (3, 2, 5). Then, find out the equation of the plane?

Solution:

The equation of the plane that crosses through A= (1,3,2) and consist of a normal vector [overrightarrow{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

Example:

In the case where the plane is crossing through the point A= (5, 6, 2) and has a normal vector [overrightarrow{n}] = (-1, 3,-7). Then, find out the equation of the plane?

Solution:

The equation of the plane that passes through the point A=(5,6,2) and has normal vector [overrightarrow{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

Passing Through Three Points

When we are given three points on a plane, we can determine the equation of the plane by solving simultaneous equations.

Suppose that ax + by + cz + d = 0 be the equation of a plane on which there are the three points: A=(1,0,2), B=(2,1,1) and C=(-1,2,1). Then, the equation of the plane is acknowledged as follows:

We already have the equation of the plane with the four unknown constants:

ax + by + cz +d = 0  ….(1)

We also obtain the following three equations by substituting the coordinates of A, B, and C into equation (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

Which provides b=3a, c=4a, d=-9a  …(2)

Substituting equation (2) into (1), we have

ax + 3ay + 4az -9a = 0

x + 3y + 4z – 9 =0

therefore, the equation of the plane crossing across the three points A=(1,0,2), B=(2,1,1), and C=(-1,2,1) is

x + 3y + 4z – 9 =0

Considering this method, we can identify the equation of a plane if we know 3 points.