Determine the shortest distance between the lines l₁ and l₂ whose vector equations are

\(\vec{r}=2 \hat{\imath}+\hat{\jmath}+\lambda(2 \hat{\imath}-\hat{\jmath}+\hat{k}) \quad \ldots(1)\)

\(\vec{r}=3 \hat{\imath}+\hat{\jmath}-\hat{k}+\mu(3 \hat{\imath}-5 \hat{\jmath}+2 \hat{k}) \quad \ldots(2)\)

Question

Question

This question was previously asked in

AAI ATC Junior Executive 25 March 2021 Official Paper (Shift 2)

Answer 1

Given:

l₁ : \(\vec{r}=2 ̂{\imath}+̂{\jmath}+\lambda(2 ̂{\imath}-̂{\jmath}+̂{k}) \quad \ldots(1)\)

l₂ : \(\vec{r}=3 ̂{\imath}+̂{\jmath}-̂{k}+\mu(3 ̂{\imath}-5 ̂{\jmath}+2 ̂{k}) \quad \ldots(2)\)

Formula:

Shortest distance between two skew lines \(\vec{r}=\vec{a_1} + \lambda \vec{b_1}\) and \(\vec{r}=\vec{a_2} + \lambda \vec{b_2}\) is given by :

\(d = |\frac{(\vec{a_2}-\vec{a_1}).(\vec{b_1}\times \vec{b_2})}{|(\vec{b_1}\times \vec{b_2})|}|\)

If \(\vec{a} = a_x̂{i} + a_ŷ{j} + a_ẑ{k}\) and \(vec{b} = b_x̂{i} + b_ŷ{j} + b_ẑ{k}\) then,

\(\vec{a}\times\vec{b} = \begin{vmatrix} {\hat{i}} &{\hat{j}} &{\hat{k}}\\a_x&a_y&a_z\\b_x&b_y&b_z\end{vmatrix}\)

Calculation:

\(\vec{a_2}-\vec{a_1}\) = î - k̂

\(\vec{b_1}\times\vec{b_2} = \begin{vmatrix} ̂{i} &̂{j} &̂{k}\\2&-1&1\\3&-5&2\end{vmatrix}\)

= 3î - ĵ - 7k̂

∴ \(|\vec{b_1}\times \vec{b_2}|\) = √(3² + (-1)² + (-7)²)

= √59

Putting all these in the distance formula,

\(d =\frac{ |( î - k̂ ).(3î -ĵ -7k̂)|}{√59} \)

⇒ d = 10/√59