Given two points A(x1, y1, z1) and B(x2, y2, z2) and a set of points (a, b, c) which represent the axis (ai + bj + ck), the task is to find the equation of plane which passes through the given points A and B and parallel to the given axis.
Input: x1 = 1, y1 = 2, z1 = 3, x2 = 3, y2 = 4, z2 = 5, a= 6, b = 7, c = 8
Output: 2x + 4y + 2z + 0 = 0
Input: x1 = 2, y1 = 3, z1 = 5, x2 = 6, y2 = 7, z2 = 8, a= 11, b = 23, c = 10.
Output: -29x + 7y + 48z + 0= 0
From the given two points on plane A and B, The directions ratios an vector equation of line AB is given by:
direction ratio = (x2 – x1, y2 – y1, z2 – z1)
Since the line is parallel to the given axis . Therefore, the cross-product of and is 0 which is given by:
d, e, and f are the coefficient of vector equation of line AB i.e.,
d = (x2 – x1),
e = (y2 – y1), and
f = (z2 – z1)
and a, b, and c are the coefficient of given axis.
The equation formed by the above determinant is given by:
Equation 1 is perpendicular to the line AB which means it is perpendicular to the required plane.
Let the Equation of the plane is given by . (Equation 2)
where A, B, and C are the direction ratio of the plane perpendicular to plane.
Since Equation 1 are Equation 2 are perpendicular to each other, therefore the value of direction ratio of Equation 1 & 2 are parallel. Then the coefficient of the plane is given by:
A = (b*f – c*e),
B = (a*f – c*d), and
C = (a*e – b*d)
Now dot product of plane and vector line AB gives the value of D as
D = -(A * d – B * e + C * f)
Below is the implementation of above approach:
-29x + 7y + 48z + 0= 0
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