{"id":6364,"date":"2021-10-13T17:52:41","date_gmt":"2021-10-13T12:22:41","guid":{"rendered":"https:\/\/mathemerize.com\/?p=6364"},"modified":"2021-10-14T22:43:17","modified_gmt":"2021-10-14T17:13:17","slug":"general-equation-of-plane-3d","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/general-equation-of-plane-3d\/","title":{"rendered":"General Equation of Plane 3D"},"content":{"rendered":"
Here you will learn what is a plane and what is the general equation of a plane.<\/p>\n
Let’s begin – <\/p>\n
A plane is a surface such that if any two points are taken on it, the line segment joining them lies completely on the surface.<\/p>\n
In other words, every point on the line segment joining any two points on a plane lies on the plane.<\/p>\n
\nThe general equation of a plane is ax + by + cz + d = 0.<\/p>\n<\/blockquote>\n
Theorem<\/strong> : Prove that every first degree equation in x, y and z represents a plane.<\/p>\n
Proof<\/strong> : Let ax + by + cz + d = 0 be a first degree in x, y and z.<\/p>\n
In order to prove that the equation ax + by + cz + d = 0 represents a plane. it is sufficient to show that every point on the line segment joining any two points on the surface represented by it lies on it. <\/p>\n
Let P(\\(x_1, y_1, z_1\\)) and Q(\\(x_2, y_2, z_2\\)) be two points on the surface represented by the equation ax + by + cz = 0. Then,<\/p>\n
\\(ax_1 + by_1 + cz_1 + d\\) = 0 ………….(i)<\/p>\n
\\(ax_2 + by_2 + cz_2 + d\\) = 0 ……………(ii)<\/p>\n
Let R be an arbitrary point on the line segment joinining P and Q. Suppose R divides PQ in the ratio \\(\\lambda\\) : 1. Then, the coordinates of R are<\/p>\n
(\\(x_1 + \\lambda x_2\\over \\lambda + 1\\), \\(y_1 + \\lambda y_2\\over \\lambda + 1\\) \\(z_1 + \\lambda z_2\\over \\lambda + 1\\)), where 0 \\(\\le\\) \\(\\lambda\\) \\(\\le\\) 1.<\/p>\n
We have to prove that R lies on the surface represented by the equation ax + by + cz + d = 0 for all values of \\(\\lambda\\) satisfying 0 \\(\\le\\) \\(\\lambda\\) \\(\\le\\) 1. For this it is sufficient to show that the coordinates of R satisfy this equation.<\/p>\n
Putting x = \\(x_1 + \\lambda x_2\\over \\lambda + 1\\), y = \\(y_1 + \\lambda y_2\\over \\lambda + 1\\) and z = \\(z_1 + \\lambda z_2\\over \\lambda + 1\\) in ax + by + cz + d = 0, we get<\/p>\n
a\\(x_1 + \\lambda x_2\\over \\lambda + 1\\) + b\\(y_1 + \\lambda y_2\\over \\lambda + 1\\) + c\\(z_1 + \\lambda z_2\\over \\lambda + 1\\) + d<\/p>\n
= \\(1\\over \\lambda + 1\\) { (\\(ax_1 + by_1 + cz_1 + d\\)) + \\(\\lambda\\) (\\(ax_2 + by_2 + cz_2 + d\\)) }<\/p>\n
= \\(1\\over \\lambda + 1\\) ( 0 + \\(\\lambda\\) 0) = 0<\/p>\n
Thus, the point R (\\(x_1 + \\lambda x_2\\over \\lambda + 1\\), \\(y_1 + \\lambda y_2\\over \\lambda + 1\\) \\(z_1 + \\lambda z_2\\over \\lambda + 1\\)) lies on the surface represented by ax + by + cz + d = 0. Since R is arbitrary point on the line segment joining P and Q. Therefore, every point on PQ lies on the surface represented by equation ax + by + cz + d = 0.<\/p>\n
Hence, the equation ax + by + cz + d = 0 represents a plane.<\/p>\n
Remark<\/strong> : To determine a plane satisfying some given condition we will have to find the values of constants a, b, c and d. It seems that are four unknowns viz a, b, c and d in the equation ax + by + cz + d = 0. But, there are only three unkowns, because the equation ax + by + cz + d = 0 can be written as<\/p>\n
(\\(a\\over d\\) x + (\\(b\\over d\\)) y + (\\(c\\over d\\)) z + 1 = 0 or,<\/p>\n
Ax + By + Cz + 1 = 0<\/p>\n
Thus, to find a plane we must have three conditions to find the values of A, B and C.<\/p>\n\n\n