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