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

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

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

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