Here you will learn the equation of plane passing through three points with example.
Let’s begin –
Equation of a Plane Passing Through a Given Point
The general equation of a plane passing through a point \((x_1, y_1, z_1\)) is
\(a(x – x_1) + b(y – y_1) + c(z – z_1)\) = 0, where a, b and c are constants.
Now, In order to find the equation of plane passing through three given points \((x_1, y_1, z_1\)), \((x_2, y_2, z_2\)) and \((x_3, y_3, z_3\)), we may use the following algorithm.
Algorithm :
1). Write the equation of a plane passing through \((x_1, y_1, z_1\)) as
\(a(x – x_1) + b(y – y_1) + c(z – z_1)\) = 0 ………..(i)
2). If the plane (i) passes through \((x_2, y_2, z_2\)) and \((x_3, y_3, z_3\)), then
\(a(x_2 – x_1) + b(y_2 – y_1) + c(z_2 – z_1)\) = 0 ………..(ii)
\(a(x_3 – x_1) + b(y_3 – y_1) + c(z_3 – z_1)\) = 0 ………..(iii)
3). Solve equation (ii) and (iii), obtained in step 2, by cross-multiplication.
4). Substituting the values of a, b, c, obtained in step 3, in equation (i) in step 1 to get the required plane.
Example : Find the equation of the plane through the points A(2, 2, -1), B(3, 4, 2) and C(7, 0 , 6).
Solution : The general equation of a plane passing through (2, 2, -1) is
a(x – 2) + b(y – 2) + c(z + 1) = 0 ……………..(i)
It will pass through B(3, 4, 2) and C(7, 0 , 6), if
a(3 – 2) + b(4 – 2) + c(2 + 1) = 0 \(\implies\) a + 2b + 3c = 0 ………..(ii)
and, a(7 – 2) + b(0 – 2) + c(6 + 1) = 0 \(\implies\) 5a – 2b + 7c = 0 …………(iii)
Solving (ii) and (iii) by cross-multiplication, we have
\(a\over 14 + 6\) = \(b\over 15 – 7\) = \(c\over -2 – 10\)
\(\implies\) \(a\over 5\) = \(b\over 2\) = \(c\over -3\) = \(\lambda\) (say)
\(\implies\) a = 5 \(\lambda\), b = 2 \(\lambda\), c = -3 \(\lambda\)
Substituting the values of a, b and c in (i), we get
5 \(\lambda\)(x – 2) + 2 \(\lambda\)(y – 2) – 3 \(\lambda\)(z + 1) = 0
\(\implies\) 5(x – 2) + 2(y – 2) – 3(z + 1) = 0
\(\implies\) 5x + 2y – 3z = 17, which is the required equation of the plane.
