Perpendicular Planes to Vectors and Points

For one particular point on the vector, however, there is only one unique plane which 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 which 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−x1) + B (y−y1) + C (z−z1) = 0, where A, B, and C are the direction ratios.

Equation of Plane Passing Through 3 Non – Collinear Points

P(x1, y1, z1), Q(x2, y2, z2), and R (x3, y3, z3) 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(x1) + b(y1) + c(z1) + d = 0

a(x2) + b(y2) + c(z2) + d = 0

a(x3) + b(y3) + c(z3) + 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.

(image will be uploaded soon)

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 = \[\frac{1}{2}\] 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) + \[\frac{1}{2}\] bz — 2b = 0

 

x – y + \[\frac{1}{2}\] z —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.