Question -
Answer -
(a) Point Plane
(0, 0, 0) 3x – 4y + 12z = 3
We know that, distanceof point P(x1, y1, z1) from the plane Ax + By+ Cz – D = 0 is given as:
Given point is (0, 0,0) and the plane is 3x – 4y + 12z = 3
= |3/√169|
= 3/13
∴ The distance is 3/13.
(b) Point Plane
(3, -2, 1) 2x – y + 2z+ 3 = 0
We know that, distanceof point P(x1, y1, z1) from the plane Ax + By+ Cz – D = 0 is given as:
Given point is (3, -2,1) and the plane is 2x – y + 2z + 3 = 0
= |13/√9|
= 13/3
∴ The distance is 13/3.
(c) Point Plane
(2, 3, -5) x + 2y – 2z= 9
We know that, distanceof point P(x1, y1, z1) from the plane Ax + By+ Cz – D = 0 is given as:
Given point is (2, 3,-5) and the plane is x + 2y – 2z = 9
= |9/√9|
= 9/3
= 3
∴ The distance is 3.
(d) Point Plane
(-6, 0, 0) 2x – 3y +6z – 2 = 0
We know that, distanceof point P(x1, y1, z1) from the plane Ax + By+ Cz – D = 0 is given as:
Given point is (-6, 0,0) and the plane is 2x – 3y + 6z – 2 = 0
= |14/√49|
= 14/7
= 2
∴ The distance is 2.