# A plane is defined by the equation: x - 7y - 18z = 0. Write the coordinates of 3 points on this plane.

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### 2 Answers

The question states that you need to find the coordinates of 3 points on the plane x - 7y - 18z = 0

Let the x and y coordinates be 0

we get 0 - 0 - 18z = 0

=> z = 0

Therefore one point that lie son the plain is ( 0, 0, 0)

Next let x = 1, y = 1

1 - 7 - 18z = 0

=> -6 - 18z = 0

=> -18z = 6

=> z = -6/18

=> z = -1/3

Another point that lies on the plain is (1, 1, -1/3)

Let x = 0 , y = 1

0 - 7 - 18z = 0

=> -18z = 7

=> z = -7 / 18

Another point that lies on the plane is (0 , 1 , -7/18)

We can in this way find any number of points that lie on the given plane. Just take any random values for two of the coordinates and calculate the third.

The coordinates of 3 points that lie on the plane are

**( 0,0,0) , (1, 1, -1/3), (0 , 1 , -7/18)**

Let z = 1/2. Then x-7y -18*(1/2) = 0. So x-7y = 9 , when z= 0.5.

Therefore x= 2, y = -1 and z = 1/2 is on x-7y-18z as 7-(-2)-18*(1/2) = 0. So **(2, -1, 1/2)** is on the plane.

Let z = 1, then x-7y = 18. So x= 4, 7 = -2 satisfies x-7y = 18, as 4 -7*-2 = 18. So **( 4,-2, 1) **is on the plane x-7y -18z = 0.

Let x = 11 , y = -1, then x-7y -18z = 0, for 11-7*-1 = 18z. Or 18z = 18. Or z = 1. So **(11, -1 , 1**) is on the plane x-7y-18z.

So the 3 points on the plane x-7y-18z = 0 are

**(2, -1, 1/2) , (4, -2, 1) , (11, -1, 1).**