Angle Between a Line and a Plane

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 \(sin\theta\) = \(al + bm + cn\over \sqrt{a^2 + b^2 + c^2}\sqrt{l^2 + m^2 + n^2}\)

= \(6 + 3 – 12\over \sqrt{4 + 1 + 9}\sqrt{9 + 4 + 16}\) = \(-4\over \sqrt{406}\)

\(\implies\) \(\theta\) = \(sin^{-1}({-4\over \sqrt{406}})\)

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