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

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

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