Here you will learn how to find angle between two planes formula with examples.
Let’s begin –
Angle Between Two Planes Formula
The angle between two planes is defined as the angle between their normals.
(a) Vector Form
The 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
\(cos\theta\) = \(\vec{n_1}.\vec{n_2}\over |\vec{n_1}||\vec{n_1}|\).
Condition of Perpendicularity : 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.
\(\vec{n_1}\).\(\vec{n_2}\) = 0
Condition of Parallelism : 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.
Therefore, there exist a scalar \(\lambda\) such that \(\vec{n_1}\) = \(\lambda\) \(\vec{n_2}\)
(b) Cartesian Form
The 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
\(cos\theta\) = \(a_1a_2 + b_1b_2 + c_1c_2\over \sqrt{{a_1}^2 + {b_1}^2 + {c_1}^2}\sqrt{{a_1}^2 + {b_1}^2 + {c_1}^2}\)
Condition of Perpendicularity : If the planes are perpendicular. Then \(\vec{n_1}\) and \(\vec{n_2}\) are perpendicular
\(a_1a_2 + b_1b_2 + c_1c_2\) = 0
Condition of Parallelism : If the lines are parallel, then \(\vec{n_1}\) and \(\vec{n_2}\) are parallel,
\(a_1\over a_2\) = \(b_1\over b_2\) = \(c_1\over c_2\)
Example : 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.
Solution : 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
\(cos\theta\) = \(\vec{n_1}.\vec{n_2}\over |\vec{n_1}||\vec{n_1}|\)
Here \(n_1\) = \(2\hat{i} – \hat{j} + \hat{k}\) and \(n_2\) = \(\hat{i} + \hat{j} + 2\hat{k}\)
\(\therefore\) \(cos\theta\) = \((2\hat{i} – \hat{j} + \hat{k}).(\hat{i} + \hat{j} + 2\hat{k})\over |2\hat{i} – \hat{j} + \hat{k}||\hat{i} + \hat{j} + 2\hat{k}|\)
= \(2 – 1 + 2\over \sqrt{4 + 1 + 1} \sqrt{1 + 1 +4}\) = \(1\over 2\)
\(\implies\) \(\theta\) = \(\pi\over 3\)
