Find the distance between the lines l_1 and l_2 given by l_1 : vector(r) = (i + 2i -4k) + λ (2i + 3j + 6k))

Question

Find the distance between the lines \(l_1\) and \(l_2\) given by

\(l_1: \vec{r} = \hat{i}+2\hat{j}-4\hat{k}+\lambda (2\hat{i}+3\hat{j}+6\hat{k});\) \(l_2: \vec{r} = 3\hat{i}+3\hat{j}-5\hat{k}+\mu (4\hat{i}+6\hat{j}+12\hat{k})\)

Answer 1

Given lines are

After observation, we get \(l_1 \big|\big| l_2\)

Therefore, it is sufficient to find the perpendicular distance of a point of line \(l_1\) to line \(l_2.\)

The coordinate of a point of \(l_1\) is P(1, 2, -4)

Also the cartesian form of line \(l_2\) is

Let Q\((\alpha, \beta, \gamma)\) be foot of perpendicular drawn from P to line \(l_2\)

\(\because\) Q(a, b, g) lie on line \(l_2\)

Again, \(\because\) \(\vec{PQ}\) is perpendicular to line \(l_2\).

\(\Rightarrow\) \(\vec{PQ}.\vec{b}=0,\) where \(\vec{b}\) is parallel vector of \(l_2\)

Therefore required perpendicular distance is