Given a point(x, y, z) in 3-D and coefficients of the equation of plane, the task is to find the mirror image of that point through the given plane.
Input: a = 1, b = -2, c = 0, d = 0, x = -1, y = 3, z = 4
Output: x3 = 1.7999999999999998, y3 = -2.5999999999999996, z3 = 4.0
Input: a = 2, b = -1, c = 1, d = 3, x = 1, y = 3, z = 4
Output: x3 = -3.0, y3 = 5.0, z3 = 2.0
Approach: Equation of plane is as ax + by + cz + d = 0. Therefore, direction ratios of the normal to the plane are (a, b, c). Let N be the foot of perpendicular from given point to the given plane so, line PN has directed ratios (a, b, c) and it passes through P(x1, y1, z1).
The equation of line PN will be as:-
(x - x1) / a = (y - y1) / b = (z - z1) / c = k
Hence any point on line PN can be written as:-
x = a*k + x1 y = b*k + y1 z = c*k + z1
since N lies in both line and plane so will satisfy(ax + by + cz + d = 0).
=>a * (a * k + x1) + b * (b * k + y1) + c * (c * k + z1) + d = 0. =>a * a * k + a * x1 + b * b * k + b * y1 + c * c * k + c * z1 + d = 0. =>(a * a + b * b + c * c)k = -a * x1 - b * y1 - c * z1 - d. =>k = (-a * x1 - b * y1 - c * z1 - d) / (a * a + b * b + c * c).
Now, the coordinates of Point N in terms of k will be:-
x2 = a * k + x1 y2 = b * k + y1 z2 = c * k + z1
Since, Point N(x2, y2, z2) is midpoint of point P(x1, y1, z1) and point Q(x3, y3, z3), coordinates of Point Q are:-
=> x3 = 2 * x2 - x1 => y3 = 2 * y2 - y1 => z3 = 2 * z2 - z1
x3 = 1.8 y3 = -2.6 z3 = 4.0
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