Here you will learn formula to find angle between a line and a plane with examples.
Let’s begin –
Angle Between a Line and a Plane
The angle between a line and a plane is the complement of the angle between the line and normal to the plane
(a) Vector Form
The angle \(\theta\) between a lines \(\vec{r}\) = \(\vec{a}\) + \(\lambda\vec{b}\) and the plane \(\vec{r}\).\(\vec{n}\) = d is given by
\(sin\theta\) = \(\vec{b}.\vec{n}\over |\vec{b}||\vec{n}|\).
Condition of Perpendicularity : 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.
\(\vec{b}\times \vec{n}\) or, \(\vec{b}\) = \(\lambda\) \(\vec{n}\) for some scalar \(\lambda\)
Condition of Parallelism : 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
\(\vec{b}\).\(\vec{n}\) = 0
(b) Cartesian Form
The 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
\(sin\theta\) = \(al + bm + cn\over \sqrt{a^2 + b^2 + c^2}\sqrt{l^2 + m^2 + n^2}\)
Condition of Perpendicularity : If the line is perpendicular to the plane, then it is parallel to its normal. Therefore,
\(l\over a\) = \(m\over b\) = \(n\over c\)
Condition of Parallelism : If the lines is parallel to the plane, then it is perpendicular to its normal. Therefore,
\(\vec{b}\).\(\vec{n}\) = 0 \(\implies\) al + bm + cn = 0
Example : 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.
Solution : Here, l = 3, m = 2 and = 4
and, a = 2, b = 1 and c = -3
So, Angle between them is \(sin\theta\) = \(al + bm + cn\over \sqrt{a^2 + b^2 + c^2}\sqrt{l^2 + m^2 + n^2}\)
= \(6 + 3 – 12\over \sqrt{4 + 1 + 9}\sqrt{9 + 4 + 16}\) = \(-4\over \sqrt{406}\)
\(\implies\) \(\theta\) = \(sin^{-1}({-4\over \sqrt{406}})\)
