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

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

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