# Image of a Point in a Plane

## Image of a Point in a Plane

Distance of a point From a Plane – Cartesian Form: Let PM be the length of the perpendicular from a point P (x₁, y₁, z₁) to the plane ax + by + cz + d = 0. Then the equation of PM is $$\frac{x-{{x}_{1}}}{a}=\frac{y-{{y}_{1}}}{b}=\frac{z-{{z}_{1}}}{c}=r$$.

The coordinates of any point on this line are (x₁ + ar, y₁ + br, z₁ + cr)

Thus, the point coincides with M if and if it lies on the plane, i.e.,

a (x₁ + ar) + b (y₁ + br) + c (z₁ + cr) + d = 0

$$r=-\frac{\left( a{{x}_{1}}+b{{y}_{1}}+c{{z}_{1}}+d \right)}{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}$$,

Now, $$PM=\sqrt{{{\left( {{x}_{1}}+ar-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{1}}+br-{{y}_{1}} \right)}^{2}}+{{\left( {{z}_{1}}+cr-{{z}_{1}} \right)}^{2}}}$$,

$$=\sqrt{\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right){{r}^{2}}}$$,

$$=\sqrt{\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)}\left| r \right|$$,

$$=\sqrt{\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)}\left| \frac{-\left( a{{x}_{1}}+b{{y}_{1}}+c{{z}_{1}}+d \right)}{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}} \right|$$,

$$=\frac{\left| -\left( a{{x}_{1}}+b{{y}_{1}}+c{{z}_{1}}+d \right) \right|}{\sqrt{\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)}}$$,

$$=\frac{\left| \left( a{{x}_{1}}+b{{y}_{1}}+c{{z}_{1}}+d \right) \right|}{\sqrt{\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)}}$$,

Also, if coordinates of M are (x₂, y₂, z₂) then

$$\frac{{{x}_{2}}-{{x}_{1}}}{a}=\frac{{{y}_{2}}-{{y}_{1}}}{b}=\frac{{{z}_{2}}-{{z}_{1}}}{c}=-\frac{\left( a{{x}_{1}}+b{{y}_{1}}+c{{z}_{1}}+d \right)}{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}$$ … (i)

Image of a Point in a Plane: Here Q is the image of P in the plane.

Therefore, M is the midpoint of PQ.

From equation (i)

$$\frac{\frac{{{x}_{3}}-{{x}_{1}}}{2}-{{x}_{1}}}{a}=\frac{\frac{{{y}_{3}}-{{y}_{1}}}{2}-{{y}_{1}}}{b}=\frac{\frac{{{z}_{3}}-{{z}_{1}}}{2}-{{z}_{1}}}{c}=-\frac{\left( a{{x}_{1}}+b{{y}_{1}}+c{{z}_{1}}+d \right)}{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}$$.

$$\frac{{{x}_{3}}-{{x}_{1}}}{a}=\frac{{{y}_{3}}-{{y}_{1}}}{b}=\frac{{{z}_{3}}-{{z}_{1}}}{c}=-\frac{2\left( a{{x}_{1}}+b{{y}_{1}}+c{{z}_{1}}+d \right)}{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}$$.

Example: A ray of light passing through the point A(1, 2, 3), strikes the plane x + y + z = 12 at B and no reflection passes through point C(3, 5, 9). Find the coordinates of point B.

Solution: $$\frac{{{x}_{3}}-{{x}_{1}}}{a}=\frac{{{y}_{3}}-{{y}_{1}}}{b}=\frac{{{z}_{3}}-{{z}_{1}}}{c}=-\frac{2\left( a{{x}_{1}}+b{{y}_{1}}+c{{z}_{1}}+d \right)}{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}$$,

Let the image of point A (1, 2,3) about x + y + z = 12 be D (p, q, r).

Then $$\frac{p-1}{1}=\frac{q-2}{1}=\frac{r-3}{1}=\frac{-2\left( 1+2+3-12 \right)}{{{1}^{2}}+{{1}^{2}}+{{1}^{2}}}$$,

p – 1 = 4

p = 5

q – 2 = 4

q = 2 + 4 = 6

r – 3 = 4

r = 3 + 4 = 7

D (p, q, r) = D (5, 6, 7)

Now equation of line CD is $$\frac{x-5}{3-5}=\frac{y-6}{5-6}=\frac{z-7}{9-7}$$.

$$\frac{x-5}{-2}=\frac{y-6}{-1}=\frac{z-7}{2}=\lambda$$.

Point B is point of intersection of AC with the plane

Let B = (-2λ + 5, -λ + 6, 2λ + 7)

This point lies on the plane.

-2λ + 5 – λ + 5 – λ + 6 + 2λ + 7 = 12

λ = 6

B = (-2 (6) + 5, – (6) + 6, 2 (6) + 7)

B = (-7, 0, 19).