Question:

Find the equation of the planes that passes through three points.

(a) (1,1,-1),(6,4,-5),(-4,-2,-3)

(b) (1,1,0),(1,2,1),(-2,2,-1).

Updated On: Sep 20, 2023
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Solution and Explanation

(a)The given points are A (1,1,-1), B (6,4,-5), and C (-4,-2,3).

\(\begin{vmatrix}1&1&-1\\6&4&-5\\-4&-2&3\end{vmatrix}\)=(12-10)-(18-20)-(-12+16)

=2+2-4 =0

Since, A, B, C are collinear points, there will be infinite number of planes passing through the given points.


(b)The given points are A (1,1,0), B (1,2,1), and C (-2,2,-1).

\(\begin{vmatrix}1&1&0\\1&2&1\\-2&-2&1\end{vmatrix}\)=(-2-2)-(2+2)
=-8≠0

Therefore, a plane will pass through the points A, B, and C.

It is known that the equation of the plane through the points, (x1,y1,z1), (x2,y2,z2), and (x3,y3,z3), is

\(\begin{vmatrix}x-x_1&y-y_1&z-z_1\\x_2-x_1&y_2-y_1&z_2-z_1\\x_3-x_1&y_3-y_1&z_3-z_1\end{vmatrix}\)=0

\(\Rightarrow \begin{vmatrix}x-1&y-1&z\\0&1&1\\-3&1&-1\end{vmatrix}\)=0

\(\Rightarrow\) (-2)(x-1)-3(y-1)+3z=0

\(\Rightarrow\)-2x-3y+3z+2+3=0

\(\Rightarrow\)- 2x-3y+3z=-5
\(\Rightarrow\) 2x+3y-3z=5

This is the cartesian equation of the required plane.

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Concepts Used:

Plane

A  surface comprising all the straight lines that join any two points lying on it is called a plane in geometry. A plane is defined through any of the following uniquely:

  • Using three non-collinear points
  • Using a point and a line not on that line
  • Using two distinct intersecting lines
  • Using two separate parallel lines

Properties of a Plane:

  • In a three-dimensional space, if there are two different planes than they are either parallel to each other or intersecting in a line.
  • A line could be parallel to a plane, intersects the plane at a single point or is existing in the plane.
  • If there are two different lines that are perpendicular to the same plane then they must be parallel to each other.
  • If there are two separate planes which are perpendicular to the same line then they must be parallel to each other.