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).