# Distance of point from plane

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

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

⇒ $$\frac{a{{x}_{1}}+b{{y}_{1}}+c{{z}_{1}}+d}{a{{x}_{2}}+b{{y}_{2}}+c{{z}_{2}}+d}<0$$

⇒ $$\frac{a{{x}_{1}}+b{{y}_{1}}+c{{z}_{1}}+d}{a{{x}_{2}}+b{{y}_{2}}+c{{z}_{2}}+d}>0$$

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

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

⇒ $$\frac{a{{x}_{1}}+b{{y}_{1}}+c{{z}_{1}}+d}{a{{x}_{2}}+b{{y}_{2}}+c{{z}_{2}}+d}<0$$

⇒ 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 $$\frac{\left| a{{x}_{1}}+b{{y}_{1}}+c{{z}_{1}}+d \right|}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}$$.

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 $$\frac{\left| {{d}_{1}}-{{d}_{2}} \right|}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}}$$.

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 $$\frac{x-{{x}_{1}}}{a}=\frac{y-{{y}_{1}}}{b}=\frac{z-{{z}_{1}}}{c}$$.

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 $$\frac{x-3}{3}=\frac{y+2}{-1}=\frac{z-1}{4}=r$$.