{"id":6380,"date":"2021-10-13T18:15:24","date_gmt":"2021-10-13T12:45:24","guid":{"rendered":"https:\/\/mathemerize.com\/?p=6380"},"modified":"2021-10-14T22:47:57","modified_gmt":"2021-10-14T17:17:57","slug":"distance-of-a-point-from-a-plane","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/distance-of-a-point-from-a-plane\/","title":{"rendered":"Distance of a Point From a Plane"},"content":{"rendered":"
Here you will learn how to find the distance of a point from a plane formula with examples.<\/p>\n
Let’s begin –<\/p>\n
The length of the perpendicular from a point having position vector \\(\\vec{a}\\) to the plane \\(\\vec{r}\\).\\(\\vec{n}\\) = d is<\/p>\n
\np = \\(|\\vec{a}.\\vec{n} – d|\\over |\\vec{n}|\\)<\/p>\n<\/blockquote>\n
(b) Cartesian Form<\/h3>\n
The length of the perpendicular form a point P\\((x_1, y_1, z_1)\\) to the plane ax + by + cz + d = 0 is given by<\/p>\n
\np = \\(|ax_1 + by_1 + cz_1 + d|\\over \\sqrt{a^2 + b^2 + c^2}\\)<\/p>\n<\/blockquote>\n
Example<\/strong><\/span> : Find the distance of the point \\(2\\hat{i} + \\hat{j} – \\hat{k}\\) from the plane \\(\\vec{r}\\).\\(\\hat{i} – 2\\hat{j} + 4\\hat{k}\\) = 9.<\/p>\n
Solution<\/strong><\/span> : We know that the perpendicular distance of a point with the position vector \\(\\vec{a}\\) from the given planes \\(\\vec{r}\\).\\(\\vec{n}\\) = d is given by<\/p>\n
p = \\(|\\vec{a}.\\vec{n} – d|\\over |\\vec{n}|\\)<\/p>\n
Here, \\(\\vec{a}\\) = \\(2\\hat{i} + \\hat{j} – \\hat{k}\\) , \\(\\vec{n}\\) = \\(\\hat{i} – 2\\hat{j} + 4\\hat{k}\\) and d = 9.<\/p>\n
So, the required distance p is given by<\/p>\n
p = \\(|(2\\hat{i} + \\hat{j} – \\hat{k}).(\\hat{i} – 2\\hat{j} + 4\\hat{k}) – 9|\\over \\sqrt{1 + 4 + 16}\\)<\/p>\n
= \\(|2 – 2 – 4 – 9|\\over \\sqrt{21}\\) = \\(13\\over \\sqrt{21}\\)<\/p>\n
Example<\/strong><\/span> : Find the distance of the point (2, 1, 0) from the plane 2x + y + 2z + 5 = 0.<\/p>\n
Solution<\/strong><\/span> : We know that the perpendicular distance of the point \\((x_1, y_1, z_1)\\) from the given planes ax+ by + cz + d = 0 is given by<\/p>\n
p = \\(|ax_1 + by_1 + cz_1 + d|\\over \\sqrt{a^2 + b^2 + c^2}\\)<\/p>\n
So, the required distance p is<\/p>\n
p = \\(|2\\times 2 + 1 + 2\\times 0 + 5|\\over \\sqrt{4 + 1 + 4}\\) = \\(10\\over 3\\)<\/p>\n\n\n