{"id":6366,"date":"2021-10-13T17:54:09","date_gmt":"2021-10-13T12:24:09","guid":{"rendered":"https:\/\/mathemerize.com\/?p=6366"},"modified":"2021-10-14T22:43:51","modified_gmt":"2021-10-14T17:13:51","slug":"equation-of-plane-passing-through-three-points","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/equation-of-plane-passing-through-three-points\/","title":{"rendered":"Equation of Plane Passing Through Three Points"},"content":{"rendered":"
Here you will learn the equation of plane passing through three points with example.<\/p>\n
Let’s begin –<\/p>\n
\nThe general equation of a plane passing through a point \\((x_1, y_1, z_1\\)) is <\/p>\n
\\(a(x – x_1) + b(y – y_1) + c(z – z_1)\\) = 0, where a, b and c are constants.<\/p>\n<\/blockquote>\n
Now, In order to find the equation of plane passing through three given points \\((x_1, y_1, z_1\\)), \\((x_2, y_2, z_2\\)) and \\((x_3, y_3, z_3\\)), we may use the following algorithm.<\/p>\n
Algorithm :<\/strong><\/p>\n
\n1)<\/strong>. Write the equation of a plane passing through \\((x_1, y_1, z_1\\)) as<\/p>\n
\\(a(x – x_1) + b(y – y_1) + c(z – z_1)\\) = 0 ………..(i)<\/p>\n
2)<\/strong>. If the plane (i) passes through \\((x_2, y_2, z_2\\)) and \\((x_3, y_3, z_3\\)), then<\/p>\n
\\(a(x_2 – x_1) + b(y_2 – y_1) + c(z_2 – z_1)\\) = 0 ………..(ii)<\/p>\n
\\(a(x_3 – x_1) + b(y_3 – y_1) + c(z_3 – z_1)\\) = 0 ………..(iii)<\/p>\n
3)<\/strong>. Solve equation (ii) and (iii), obtained in step 2, by cross-multiplication.<\/p>\n
4)<\/strong>. Substituting the values of a, b, c, obtained in step 3, in equation (i) in step 1 to get the required plane.<\/p>\n<\/blockquote>\n
Example<\/strong><\/span> : Find the equation of the plane through the points A(2, 2, -1), B(3, 4, 2) and C(7, 0 , 6).<\/p>\n
Solution<\/span><\/strong> : The general equation of a plane passing through (2, 2, -1) is<\/p>\n
a(x – 2) + b(y – 2) + c(z + 1) = 0 ……………..(i)<\/p>\n
It will pass through B(3, 4, 2) and C(7, 0 , 6), if<\/p>\n
a(3 – 2) + b(4 – 2) + c(2 + 1) = 0 \\(\\implies\\) a + 2b + 3c = 0 ………..(ii)<\/p>\n
and, a(7 – 2) + b(0 – 2) + c(6 + 1) = 0 \\(\\implies\\) 5a – 2b + 7c = 0 …………(iii)<\/p>\n
Solving (ii) and (iii) by cross-multiplication, we have<\/p>\n
\\(a\\over 14 + 6\\) = \\(b\\over 15 – 7\\) = \\(c\\over -2 – 10\\) <\/p>\n
\\(\\implies\\) \\(a\\over 5\\) = \\(b\\over 2\\) = \\(c\\over -3\\) = \\(\\lambda\\) (say)<\/p>\n
\\(\\implies\\) a = 5 \\(\\lambda\\), b = 2 \\(\\lambda\\), c = -3 \\(\\lambda\\)<\/p>\n
Substituting the values of a, b and c in (i), we get<\/p>\n
5 \\(\\lambda\\)(x – 2) + 2 \\(\\lambda\\)(y – 2) – 3 \\(\\lambda\\)(z + 1) = 0<\/p>\n
\\(\\implies\\) 5(x – 2) + 2(y – 2) – 3(z + 1) = 0<\/p>\n
\\(\\implies\\) 5x + 2y – 3z = 17, which is the required equation of the plane.<\/p>\n\n\n