{"id":6372,"date":"2021-10-13T18:03:28","date_gmt":"2021-10-13T12:33:28","guid":{"rendered":"https:\/\/mathemerize.com\/?p=6372"},"modified":"2021-10-14T22:45:49","modified_gmt":"2021-10-14T17:15:49","slug":"equation-of-plane-in-normal-form","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/equation-of-plane-in-normal-form\/","title":{"rendered":"Equation of Plane in Normal Form"},"content":{"rendered":"
Here you will learn equation of plane in normal form with example.<\/p>\n
Let’s begin –<\/p>\n
\nThe vector equation of a plane normal to unit vector \\(\\hat{n}\\) and at a distance d from the origin is<\/p>\n
\\(\\vec{r}.\\hat{n}\\) = d<\/p>\n<\/blockquote>\n
Remark 1 :<\/strong> The vector equation of ON is \\(\\vec{r}\\) = \\(\\vec{0}\\) + \\(\\lambda\\) \\(\\hat{n}\\) and the position vector of N is d\\(\\hat{n}\\) as it is at a distance d from the origin from the origin O.<\/p>\n
(b) Cartesian Form <\/h3>\n
\nIf l, m, n are direction cosines of the normal to a given plane which is at a distance p from the origin, then the equation of the plane is<\/p>\n
lx + my + nz = p<\/p>\n<\/blockquote>\n
Note<\/strong> : The equation \\(\\vec{r}.\\vec{n}\\) = d is in normal form if \\(\\vec{n}\\) is a unit vector and in such a case d on the right hand side denotes the distance of the plane from the origin. If \\(\\vec{n}\\) is not a unit vector, then to reduce the equation \\(\\vec{r}.\\vec{n}\\) = d to normal form divide both sides by | \\(\\vec{n}\\) | to obtain<\/p>\n
\\(\\vec{r}\\).\\(\\vec{n}\\over |\\vec{n}|\\) = \\(d\\over |\\vec{n}|\\) \\(\\implies\\) \\(\\vec{r}.\\hat{n}\\) = \\(d\\over |\\vec{n}|\\)<\/p>\n
Example<\/strong><\/span> : Find the vector equation of a plane which is at a distance of 8 units from the origin and which is normal to the vector \\(2\\hat{i} + \\hat{j} + 2\\hat{k}\\).<\/p>\n
Solution<\/span><\/strong> : Here, d = 8 and \\(\\vec{n}\\) = \\(2\\hat{i} + \\hat{j} + 2\\hat{k}\\)<\/p>\n
\\(\\therefore\\) \\(\\hat{n}\\) = \\(\\vec{n}\\over |\\vec{n}|\\) = \\(2\\hat{i} + \\hat{j} + 2\\hat{k}\\over \\sqrt{4 + 1 + 4}\\)<\/p>\n
= \\({2\\over 3}\\hat{i} + {1\\over 3}\\hat{j} + {2\\over 3}\\hat{k}\\)<\/p>\n
Hence, the required equation of the plane is<\/p>\n
\\(\\vec{r}\\).(\\({2\\over 3}\\hat{i} + {1\\over 3}\\hat{j} + {2\\over 3}\\hat{k}\\)) = 8 <\/p>\n
[ By using \\(\\vec{r}.\\hat{n}\\) = d ]<\/p>\n
or, \\(\\vec{r}\\).(\\(2\\hat{i} + \\hat{j} + 2\\hat{k}\\)) = 24<\/p>\n\n\n