{"id":6390,"date":"2021-10-13T18:23:47","date_gmt":"2021-10-13T12:53:47","guid":{"rendered":"https:\/\/mathemerize.com\/?p=6390"},"modified":"2022-01-16T17:05:29","modified_gmt":"2022-01-16T11:35:29","slug":"image-of-a-point-in-a-plane","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/image-of-a-point-in-a-plane\/","title":{"rendered":"Image of a Point in a Plane – Formula and Example"},"content":{"rendered":"
Here you will learn how to find image of a point in a plane formula with examples.<\/p>\n
Let’s begin –<\/p>\n
Let P and Q be two points and let \\(\\pi\\) be a plane such that<\/p>\n
(i) line PQ is perpendicular to the plane \\(\\pi\\) and,<\/p>\n
(ii) mid-point of PQ lies on the plane \\(\\pi\\).<\/p>\n
Then, either of the point is the image of the other in the plane \\(\\pi\\).<\/p>\n
In order to find the image of a points \\((x_1, y_1, z_1)\\) in the plane ax + by + cz + d = 0, we may use the following algorithm.<\/p>\n
Algorithm : <\/strong><\/p>\n 1). Write the equations of the line passing through P and normal to the given plane as \\(x – x_1\\over a\\) = \\(y – y_1\\over b\\) = \\(z – z_1\\over c\\).<\/p>\n 2). Write the coordinates of image Q as \\((x_1 + ar, y_1 + br, z_1 + cr)\\).<\/p>\n 3). Find the coordinates of the mid-point R of PQ.<\/p>\n 4). Obtain the value of r by substituting the coordinates of R in the equation of the plane.<\/p>\n 5). Put the value of r in the coordinates of Q.<\/p>\n<\/blockquote>\n Example<\/span><\/strong> : Find the image of the point P (3, -2, 1) in the plane 3x – y + 4z = 2.<\/p>\n Solution<\/span><\/strong> : Let Q be the image of P(3, -2, 1) in the plane 3x – y + 4z = 2. Then PQ is normal to the plane. Therefore, direction ratios of PQ are proportional to 3, -1, 4. Since PQ passes through P(3, -2, 1) and has direction ratios proportional to 3, -1, 4.<\/p>\n Therefore, equation of PQ is<\/p>\n \\(x – 3\\over 3\\) = \\(y + 2\\over -1\\) = \\(z – 1\\over 4\\) = r (say)<\/p>\n Let the coordinates of Q be (3r + 3, -r – 2, 4r + 1). Let R be the mid-point of PQ. Then, R lies on the plane 3x – y + 4z = 2. <\/p>\n The coordinates of R are (\\({3r + 3 + 3\\over 3}, {-r – 2 – 2\\over 2}, {4r + 1 +1\\over 2}\\))<\/p>\n or, (\\({3r + 6\\over 3}, {-r – 4\\over 2}, {2r + 1}\\))<\/p>\n Since r lies on the plane 3x – y + 4z = 2.<\/p>\n \\(3({3r + 6\\over 2}) – ({-r – 4\\over 2})\\) + 4(2r + 1) = 2<\/p>\n \\(\\implies\\) 13r = -13 \\(\\implies\\) r = – 1<\/p>\n Putting r = -1 in (3r + 3, -r – 2, 4r + 1), we obtain the coordinates of Q as (0, – 1, -3).<\/p>\n Hence, the image of (3, -2, 1) in the plane 3x – y + 4z = 2 is (0, -1, -3).<\/p>\n\n\n\n