{"id":6374,"date":"2021-10-13T18:07:36","date_gmt":"2021-10-13T12:37:36","guid":{"rendered":"https:\/\/mathemerize.com\/?p=6374"},"modified":"2021-10-14T22:46:15","modified_gmt":"2021-10-14T17:16:15","slug":"angle-between-two-planes-formula","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/angle-between-two-planes-formula\/","title":{"rendered":"Angle Between Two Planes Formula"},"content":{"rendered":"
Here you will learn how to find angle between two planes formula with examples.<\/p>\n
Let’s begin –<\/p>\n
The angle between two planes is defined as the angle between their normals.<\/p>\n
\nThe angle \\(\\theta\\) between the planes \\(\\vec{r}\\).\\(\\vec{n_1}\\) = \\(\\vec{d_1}\\) and \\(\\vec{r}\\).\\(\\vec{n_2}\\) = \\(\\vec{d_2}\\) is given by<\/p>\n
\\(cos\\theta\\) = \\(\\vec{n_1}.\\vec{n_2}\\over |\\vec{n_1}||\\vec{n_1}|\\).<\/p>\n<\/blockquote>\n
Condition of Perpendicularity<\/strong> : If the planes \\(\\vec{r}\\).\\(\\vec{n_1}\\) = \\(\\vec{d_1}\\) and \\(\\vec{r}\\).\\(\\vec{n_2}\\) = \\(\\vec{d_2}\\) are perpendicular, then \\(\\vec{n_1}\\) and \\(\\vec{n_2}\\) are perpendicular.<\/p>\n
\n\\(\\vec{n_1}\\).\\(\\vec{n_2}\\) = 0<\/p>\n<\/blockquote>\n
Condition of Parallelism<\/strong> : If the planes \\(\\vec{r}\\).\\(\\vec{n_1}\\) = \\(\\vec{d_1}\\) and \\(\\vec{r}\\).\\(\\vec{n_2}\\) = \\(\\vec{d_2}\\) are parallel, then \\(\\vec{n_1}\\) and \\(\\vec{n_2}\\) are parallel.<\/p>\n
\nTherefore, there exist a scalar \\(\\lambda\\) such that \\(\\vec{n_1}\\) = \\(\\lambda\\) \\(\\vec{n_2}\\)<\/p>\n<\/blockquote>\n
(b) Cartesian Form<\/h3>\n
\nThe angle \\(\\theta\\) between the planes \\(a_1x + b_1y + c_1z + d_1\\) = 0 and \\(a_2x + b_2y + c_2z + d_2\\) = 0 is given by<\/p>\n
\\(cos\\theta\\) = \\(a_1a_2 + b_1b_2 + c_1c_2\\over \\sqrt{{a_1}^2 + {b_1}^2 + {c_1}^2}\\sqrt{{a_1}^2 + {b_1}^2 + {c_1}^2}\\)<\/p>\n<\/blockquote>\n
Condition of Perpendicularity<\/strong> : If the planes are perpendicular. Then \\(\\vec{n_1}\\) and \\(\\vec{n_2}\\) are perpendicular<\/p>\n
\n\\(a_1a_2 + b_1b_2 + c_1c_2\\) = 0<\/p>\n<\/blockquote>\n
Condition of Parallelism<\/strong> : If the lines are parallel, then \\(\\vec{n_1}\\) and \\(\\vec{n_2}\\) are parallel, <\/p>\n
\n\\(a_1\\over a_2\\) = \\(b_1\\over b_2\\) = \\(c_1\\over c_2\\)<\/p>\n<\/blockquote>\n
Example<\/strong><\/span> : Find the angle between the planes \\(\\vec{r}\\).(\\(2\\hat{i} – \\hat{j} + \\hat{k}\\)) = 6 and \\(\\vec{r}\\).(\\(\\hat{i} + \\hat{j} + 2\\hat{k}\\)) = 5.<\/p>\n
Solution<\/span><\/strong> : We know that the angle between the planes \\(\\vec{r}\\).\\(\\vec{n_1}\\) = \\(\\vec{d_1}\\) and \\(\\vec{r}\\).\\(\\vec{n_2}\\) = \\(\\vec{d_2}\\) is given by<\/p>\n
\\(cos\\theta\\) = \\(\\vec{n_1}.\\vec{n_2}\\over |\\vec{n_1}||\\vec{n_1}|\\)<\/p>\n
Here \\(n_1\\) = \\(2\\hat{i} – \\hat{j} + \\hat{k}\\) and \\(n_2\\) = \\(\\hat{i} + \\hat{j} + 2\\hat{k}\\)<\/p>\n
\\(\\therefore\\) \\(cos\\theta\\) = \\((2\\hat{i} – \\hat{j} + \\hat{k}).(\\hat{i} + \\hat{j} + 2\\hat{k})\\over |2\\hat{i} – \\hat{j} + \\hat{k}||\\hat{i} + \\hat{j} + 2\\hat{k}|\\)<\/p>\n
= \\(2 – 1 + 2\\over \\sqrt{4 + 1 + 1} \\sqrt{1 + 1 +4}\\) = \\(1\\over 2\\)<\/p>\n
\\(\\implies\\) \\(\\theta\\) = \\(\\pi\\over 3\\)<\/p>\n\n\n