Question

Find the equations 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)

Answer 1

Consider three points lying on a plane,

A( x 1 , y 1 , z 1 ), B( x 2 , y 2 , z 2 ), C( x 3 , y 3 , z 3 )

To compute the equation of plane, first we will check the determinant of the given point. If the determinant is equal to 0, then infinite number of planes can be passed and if the determinant is not equal to 0, then the equation of plane can be formed by the given formula,

| 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 |=0(1)

(a)

The three given points are A( 1,1,−1 ),B( 6,4,−5 ),C( −4,−2,3 ).

Calculating the determinant,

| 1 1 −1 6 4 −5 −4 −2 3 |=1( 12−10 )−1( 18−20 )−1( −12+16 ) =2+2−4 =0

Since the determinant is equal to 0; therefore, infinite number of planes can be passed.

(b)

The three given points are A( 1,1,0 ),B( 1,2,1 ),C( −2,2,−1 ).

Calculating the determinant,

| 1 1 0 1 2 1 −2 2 −1 |=1( −2−2 )−1( 2+2 )−0 =−8 ≠0

Since the determinant is not equal to 0; therefore, the equation of plane can be obtained from equation (1).

Compare the points,

x 1 =1

x 2 =1

x 3 =−2

y 1 =1

y 2 =2

y 3 =2

z 1 =0

z 2 =1

z 3 =−1

Substitute the values in equation (8).

| x−1 y−1 z−0 0 1 1 −3 1 −1 |=0 −2x−3y+3z+2+3=0 −2x−3y+3z=−5 2x+3y−3z=5

Thus, the required equation of plane is 2x+3y−3z=5.