{"id":6382,"date":"2021-10-13T18:17:15","date_gmt":"2021-10-13T12:47:15","guid":{"rendered":"https:\/\/mathemerize.com\/?p=6382"},"modified":"2021-10-14T22:48:23","modified_gmt":"2021-10-14T17:18:23","slug":"distance-between-parallel-planes","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/distance-between-parallel-planes\/","title":{"rendered":"Distance Between Parallel Planes"},"content":{"rendered":"
Here you will learn how to find distance between parallel planes with examples.<\/p>\n
Let’s begin –<\/p>\n
Let ax + by + cz + \\(d_1\\) = 0 and ax + by + cz + \\(d_2\\) = 0 be two parallel planes. In order to find the distance between them, we may follow the following algorithm.<\/p>\n
Algorithm :<\/strong><\/p>\n 1).<\/strong> Take an aribitrary point P\\((x_1, y_1, z_1)\\) on one of the planes, say ax + by + cz + \\(d_1\\) = 0.<\/p>\n 2).<\/strong> Find the length of the perpendicular ‘d’ drawn form P \\((x_1, y_1, z_1)\\) on the other plane i.e ax + by + cz + \\(d_2\\) = 0. Clearly,<\/p>\n d = |\\(ax_1 + by_1 + cz_1 + d_2\\over \\sqrt{a^2 + b^2 + c^2}\\)|<\/p>\n 3).<\/strong> As P\\((x_1, y_1, z_1)\\) lies on the plane ax + by + cz + \\(d_1\\) = 0.<\/p>\n \\(\\therefore\\) \\(ax_1 + by_1 + cz_1 + d_1\\) = 0 \\(\\implies\\) \\(ax_1 + by_1 + cz_1\\) = \\(-d_1\\)<\/p>\n 4).<\/strong> Substitute \\(ax_1 + by_1 + cz_1\\) = \\(-d_1\\) in the expression for d obtained in step 2 to get d = \\(|d_2 – d_1|\\over \\sqrt{a^2 + b^2 + c^2}\\), which gives the required distance.<\/p>\n<\/blockquote>\n Remark 1<\/strong> : So the formula used to find the distance between the parallel planes ax + by + cz + \\(d_1\\) = 0 and ax + by + cz + \\(d_2\\) = 0 is<\/p>\n d = \\(|d_1 – d_2|\\over \\sqrt{a^2 + b^2 + c^2}\\)<\/p>\n<\/blockquote>\n Remark 2<\/strong> : The distance between the parallel planes ax + by + cz + \\(d_1\\) = 0 and \\(\\lambda\\)(ax + by + cz) + \\(d_2\\) = 0 is given by<\/p>\n d = \\(|d_1 – {d_2\/\\lambda}|\\over \\sqrt{a^2 + b^2 + c^2}\\)<\/p>\n<\/blockquote>\n Example<\/span><\/strong> : Find the distance between the parallal planes x + y – z + 4 = 0 and x + y – z + 5 = 0.<\/p>\n Solution<\/strong><\/span> : Here, \\(d_1\\) = 4 and \\(d_2\\) = 5<\/p>\n So, d = \\(|d_1 – d_2|\\over \\sqrt{a^2 + b^2 + c^2}\\)<\/p>\n = \\(|4 – 5|\\over \\sqrt{3}\\) = \\(1\\over \\sqrt{3}\\)<\/p>\n\n\n\n
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