# Equation of Plane Passing Through Three Points

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.