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

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