Angle Between Two Planes Formula

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 $n_2$ = $\hat{i} + \hat{j} + 2\hat{k}$

$\therefore$  $cos\theta$ = $(2\hat{i} – \hat{j} + \hat{k}).(\hat{i} + \hat{j} + 2\hat{k})\over |2\hat{i} – \hat{j} + \hat{k}||\hat{i} + \hat{j} + 2\hat{k}|$

= $2 – 1 + 2\over \sqrt{4 + 1 + 1} \sqrt{1 + 1 +4}$ = $1\over 2$

$\implies$ $\theta$ = $\pi\over 3$