# Angle Between Two Lines in 3d

Here you learn formula for angle between two lines in 3d in both vector form and cartesian form with examples.

Let’s begin –

## Angle Between Two Lines in 3d

### (a) Vector Form

Let the vector equations of the two lines be $$\vec{r}$$ = $$\vec{a_1}$$ + $$\lambda \vec{b_1}$$ and $$\vec{r}$$ = $$\vec{a_2}$$ + $$\mu \vec{b_2}$$.

These two lines are parallel to the vectors $$\vec{b_1}$$ and $$\vec{b_2}$$ respectively. Therefore, angle between  these two lines is equal to the angle between $$\vec{b_1}$$ and $$\vec{b_2}$$.

Thus, if $$\theta$$ is the angle between the given lines, then

$$cos\theta$$ = $$\vec{b_1}.\vec{b_2}\over |\vec{b_1}||\vec{b_2}|$$

Condition of Perpendicularity : If the lines $$\vec{b_1}$$ and $$\vec{b_2}$$ are perpendicular. Then

$$\vec{b_1}$$. $$\vec{b_2}$$ = 0

Condition of Parallelism : If the lines are parallel, then $$\vec{b_1}$$ and $$\vec{b_2}$$ are parallel,

$$\therefore$$ $$\vec{b_1}$$ = $$\lambda \vec{b_2}$$ for some scalar $$\lambda$$

### (b) Cartesian Form

Let the cartesian equation of the two lines be

$$x – x_1\over a_1$$ = $$y – y_1\over b_1$$ = $$z – z_1\over c_1$$                       …………(i)

and $$x – x_1\over a_2$$ = $$y – y_1\over b_2$$ = $$z – z_1\over c_2$$                 …………(ii)

Direction ratios of line (i) are proportional to $$a_1$$, $$b_1$$, $$c_1$$.

$$\therefore$$ $$\vec{m_1}$$ = Vector parallel to line (i) = $$a_1\hat{i} + b_1\hat{j} + c_1\hat{k}$$.

Direction ratios of line (ii) are proportional to $$a_2$$, $$b_2$$, $$c_2$$.

$$\therefore$$ $$\vec{m_2}$$ = Vector parallel to line (ii) = $$a_2\hat{i} + b_2\hat{j} + c_2\hat{k}$$.

Let $$\theta$$ be the angle between (i) and (ii).

Then, $$\theta$$ is also the angle between $$\vec{m_1}$$ and $$\vec{m_2}$$.

$$\therefore$$ $$cos\theta$$ = $$\vec{m_1}.\vec{m_2}\over |\vec{m_1}||\vec{m_2}|$$

$$\implies$$ $$cos\theta$$ = $$a_1a_2 + b_1b_2 + c_1c_2\over \sqrt{{a_1}^2 + {b_1}^2 + {c_1}^2}\sqrt{{a_1}^2 + {b_1}^2 + {c_1}^2}$$

Condition of Perpendicularity : If the lines are perpendicular. Then

$$\vec{m_1}$$. $$\vec{m_2}$$ = 0 $$\implies$$ $$a_1a_2 + b_1b_2 + c_1c_2$$ = 0

Condition of Parallelism : If the lines are parallel, then $$\vec{m_1}$$ and $$\vec{m_2}$$ are parallel,

$$\therefore$$ $$\vec{m_1}$$ = $$\lambda \vec{m_2}$$ for some scalar $$\lambda$$

$$\implies$$ $$a_1\over a_2$$ = $$b_1\over b_2$$ = $$c_1\over c_2$$

Example : Find the angle between the lines

$$\vec{r}$$ = $$3\hat{i} + 2\hat{j} – 4\hat{k}$$ + $$\lambda$$ ($$\hat{i} + 2\hat{j} + 2\hat{k}$$) and

$$\vec{r}$$ = ($$5\hat{j} – 2\hat{k}$$) + $$\mu$$ ($$3\hat{i} + 2\hat{j} + 6\hat{k}$$)

Solution : Let $$\theta$$ be the angle between the given lines. These given lines are parallel to the vectors $$\vec{b_1}$$ = $$\hat{i} + 2\hat{j} + 2\hat{k}$$ and $$\vec{b_2}$$ = $$3\hat{i} + 2\hat{j} + 6\hat{k}$$ respectively.

So, the angle $$\theta$$ between them is given by

$$cos\theta$$ = $$\vec{b_1}.\vec{b_2}\over |\vec{b_1}||\vec{b_2}|$$

$$\implies$$ $$cos\theta$$ = $$3 + 4 + 12\over \sqrt{1 + 4 + 4}\sqrt{9 + 4 + 36}$$ = $$19\over 21$$

Hence, $$\theta$$ = $$cos^{-1}({19\over 21})$$