Distance of point from plane

Let the equation of a plane be ax + by + cz + d = 0

1. When P and Q are on the same side of the plane

=> (ax₁ + by₁ + cz₁ + d)/ (ax₂ +by₂ + cz₂ + d) < 0

=> (ax₁ + by₁ + cz₁ + d)/ (ax₂ +by₂ + cz₂ + d) > 0

=> ax₁ + by₁ + cz₁ + d and ax₂ +by₂ + cz₂ + d are of the same sign.

2. When P and Q are on the opposite side of the plane

=> (ax₁ + by₁ + cz₁ + d)/ (ax₂ +by₂ + cz₂ + d) < 0

=> ax₁ + by₁ +cz₁ + d and ax₂ + by₂ + cz₂+ d are of opposite sign.

Perpendicular Distance of a point from a plane: The length of the perpendicular from the point P (x₁, y₁, z₁) on the plane ax + by + cz + d = 0 is |ax₁ + by₁ + cz₁ + d|/ √ (a₁² + b₁² + c₁²)

Distance between Parallel Planes:

Let ax₁ + by₁ + cz₁ + d =0 … (i)

And, ax₂ + by₂ + cz₂ + d =0 … (ii)

Be two parallel planes

Then, the distance between them |d₁ - d₂|/ √ (a₁² + b₁² + c₁²)

Image of a Point in a Plane: Let P an Q be two points and let π be a plane such that

i) Line PQ is perpendicular to the plane π, and

ii) Mid-point of PQ lies on the plane π. Then either of the point is the image of the other in the plane π.

In order to find the image of a point in a given plane. We may use the following algorithm

Algorithm:

Step I: Write the equations of the line passing through P and normal - to the given plane as (x - x₁)/ a = (y - y₁)/ b = (z - z₁)/ c.

Step II: Write the coordinates of image Q as (x₁ + ar, y₁ + br, z₁ + cr).

Step III: Find the coordinates of the mid-point R of PQ.

Step IV: Obtain the value of r by putting the coordinates of R in the equation of the plane.

Step V: Put the value of r in the coordinates of Q.

Example: Find the image of the point (3, -2, 1) in the plane 3x – y + 4z = 2.

Solution: Let Q be the image of the point P (3, -2, 1) in the plane 3x – y + 4z = 2. Then, PQ is normal to the plane.

Therefore, direction ratios of PQ are 3, -1, 4. Since PQ passes through P (3, -2, 1) and has direction ratios 3, -1, 4.

Therefore, its equations is (x - 3)/ 3 = (y + 2)/ -1 = (z – 1)/ 4 = r.