{"id":6386,"date":"2021-10-13T18:20:17","date_gmt":"2021-10-13T12:50:17","guid":{"rendered":"https:\/\/mathemerize.com\/?p=6386"},"modified":"2021-10-14T22:49:26","modified_gmt":"2021-10-14T17:19:26","slug":"intersection-of-a-line-and-a-plane","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/intersection-of-a-line-and-a-plane\/","title":{"rendered":"Intersection of a Line and a Plane"},"content":{"rendered":"
Here you will learn how to find intersection of a line and a plane with examples.<\/p>\n
Let’s begin – <\/p>\n
Let the equation of a line be \\(x – x_1\\over l\\) = \\(y – y_1\\over m\\) = \\(z – z_1\\over n\\) and that of a plane be ax + by + cz + d = 0.<\/p>\n
The coordinates of any point on the line \\(x – x_1\\over l\\) = \\(y – y_1\\over m\\) = \\(z – z_1\\over n\\) is given <\/p>\n
\\(x – x_1\\over l\\) = \\(y – y_1\\over m\\) = \\(z – z_1\\over n\\) = r (say)<\/p>\n
or, \\((x_1 + lr, y_1 + mr, z + nr)\\) …………(i)<\/p>\n
If it lies on the plane ax + by + cz + d = 0, then<\/p>\n
\\(a(x_1 + lr)\\) + \\(b(y_1 + mr)\\) + \\(c(z_1 + nr)\\) + d = 0<\/p>\n
\\(\\implies\\) \\((ax_1 + by_1 + cz_1 + d)\\) + r(al + bm + cn) = 0<\/p>\n
\\(\\implies\\) r = -\\((ax_1 + by_1 + cz_1 + d)\\over al + bm + cn\\)<\/p>\n
Substituting the value of r in (i), we obtain the coordinates of the required point of intersection.<\/p>\n
In order to find the coordinates of the point of intersection of a line and a plane, we may use the following algorithm,<\/p>\n
Algorithm :<\/strong><\/p>\n 1). Write the coordinates of any point on the line in terms of some parameters r (say).<\/p>\n 2). Substitute these coordinates in the equation of the plane to obtain the value of r.<\/p>\n 3). Put the value of r in the coordinates of the point in step 1.<\/p>\n<\/blockquote>\n Example<\/strong><\/span> : Find the coordinates of the point where the line through the points A(3, 4, 1) and B(5, 1, 6) crosses the XY-plane.<\/p>\n Solution<\/span><\/strong> : The equation of the line passing through A and B is<\/p>\n \\(x – 3\\over 5 – 3\\) = \\(y – 4\\over 1 – 4\\) = \\(z – 1\\over 6 – 1\\)<\/p>\n or, \\(x – 3\\over 2\\) = \\(y – 4\\over -3\\) = \\(z – 1\\over 5\\)<\/p>\n The coordinates of any point on this line are given by<\/p>\n \\(x – 3\\over 2\\) = \\(y – 4\\over -3\\) = \\(z – 1\\over 5\\) = \\(\\lambda\\) <\/p>\n \\(\\implies\\) x = \\(2\\lambda + 3\\), y = \\(-3\\lambda + 4\\), z = \\(5\\lambda + 1\\)<\/p>\n So, \\((2\\lambda + 3, -3\\lambda + 4, 5\\lambda + 1)\\) are coordinates of any point on the line passing through A and B. If it lies on XY-plane i.e z = 0.Then.<\/p>\n \\(5\\lambda + 1\\) = 0 \\(\\implies\\) \\(\\lambda\\) = -\\(1\\over 5\\)<\/p>\n Thus, the coordinates of the required point are (13\/5, 23\/5, 0).<\/p>\n\n\n\n