{"id":6384,"date":"2021-10-13T18:18:42","date_gmt":"2021-10-13T12:48:42","guid":{"rendered":"https:\/\/mathemerize.com\/?p=6384"},"modified":"2021-10-14T22:49:05","modified_gmt":"2021-10-14T17:19:05","slug":"angle-between-a-line-and-a-plane","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/angle-between-a-line-and-a-plane\/","title":{"rendered":"Angle Between a Line and a Plane"},"content":{"rendered":"
Here you will learn formula to find angle between a line and a plane with examples.<\/p>\n
Let’s begin –<\/p>\n
The angle between a line and a plane is the complement of the angle between the line and normal to the plane<\/p>\n
\nThe angle \\(\\theta\\) between a lines \\(\\vec{r}\\) = \\(\\vec{a}\\) + \\(\\lambda\\vec{b}\\) and the plane \\(\\vec{r}\\).\\(\\vec{n}\\) = d is given by<\/p>\n
\\(sin\\theta\\) = \\(\\vec{b}.\\vec{n}\\over |\\vec{b}||\\vec{n}|\\).<\/p>\n<\/blockquote>\n
Condition of Perpendicularity<\/strong> : If the line is perpendicular to the plane, then it is parallel to the normal to the plane. Therefore, \\(\\vec{b}\\) and \\(\\vec{n}\\) are parallel.<\/p>\n
\n\\(\\vec{b}\\times \\vec{n}\\) or, \\(\\vec{b}\\) = \\(\\lambda\\) \\(\\vec{n}\\) for some scalar \\(\\lambda\\)<\/p>\n<\/blockquote>\n
Condition of Parallelism<\/strong> : If the line is parallel to the plane, then it is perpendicular to the normal to the plane. Therefore, \\(\\vec{b}\\) and \\(\\vec{n}\\) are perpendicular<\/p>\n
\n\\(\\vec{b}\\).\\(\\vec{n}\\) = 0<\/p>\n<\/blockquote>\n
(b) Cartesian Form<\/h3>\n
\nThe angle \\(\\theta\\) between the lines \\(x – x1\\over l\\) = \\(y – y1\\over m\\) = \\(z – z1\\over n\\) and the plane ax + by + cz + d = 0 is given by<\/p>\n
\\(sin\\theta\\) = \\(al + bm + cn\\over \\sqrt{a^2 + b^2 + c^2}\\sqrt{l^2 + m^2 + n^2}\\)<\/p>\n<\/blockquote>\n
Condition of Perpendicularity<\/strong> : If the line is perpendicular to the plane, then it is parallel to its normal. Therefore,<\/p>\n
\n\\(l\\over a\\) = \\(m\\over b\\) = \\(n\\over c\\)<\/p>\n<\/blockquote>\n
Condition of Parallelism<\/strong> : If the lines is parallel to the plane, then it is perpendicular to its normal. Therefore,<\/p>\n
\n\\(\\vec{b}\\).\\(\\vec{n}\\) = 0 \\(\\implies\\) al + bm + cn = 0<\/p>\n<\/blockquote>\n
Example<\/strong><\/span> : Find the angle between the line \\(x + 1\\over 3\\) = \\(y – 1\\over 2\\) = \\(z – 2\\over 4\\) and the plane 2x + y – 3z + 4 = 0.<\/p>\n
Solution<\/span><\/strong> : Here, l = 3, m = 2 and = 4<\/p>\n
and, a = 2, b = 1 and c = -3<\/p>\n
So, Angle between them is \\(sin\\theta\\) = \\(al + bm + cn\\over \\sqrt{a^2 + b^2 + c^2}\\sqrt{l^2 + m^2 + n^2}\\)<\/p>\n
= \\(6 + 3 – 12\\over \\sqrt{4 + 1 + 9}\\sqrt{9 + 4 + 16}\\) = \\(-4\\over \\sqrt{406}\\)<\/p>\n
\\(\\implies\\) \\(\\theta\\) = \\(sin^{-1}({-4\\over \\sqrt{406}})\\)<\/p>\n\n\n