The given lines are non-parallel lines. There is a unique line segment PQ (P lying on one and Q on the other) at right angles to both lines. PQ is the shortest distance between the lines. Hence, the shortest possible distance between the insects = PQ
The position vector of P lying on the line
\(\vec r = 6 \hat i + 2\hat j + 2\hat k + \lambda (\hat i - 2\hat j + 2\hat k)\) is \((6 + \lambda )\hat i + (2 - 2\lambda)\hat j + (2 + 2\lambda)\hat k\) for some λ
The position vector of Q lying on the line
\(\vec r= -4\hat i - \hat k + \mu (3\hat i - 2\hat j - 2\hat k)\) is \((-4 + 3\mu )\hat i + (-2\mu )\hat j + (-1 - 2\mu )\hat k\) for some μ
\(\vec {PQ} = (-10 + 3\mu - \lambda)\hat i + (-2\mu - 2 + 2\lambda)\hat j + (-3 - 2\mu -2\lambda )\hat k\)
Since PQ is perpendicular to both lines
\((-10 + 3\mu - \lambda) + (-2\mu - 2 + 2\lambda)(-2) + (-3-2\mu -2\lambda)2 = 0,\)
i.e., \(μ - 3λ = 4\) ...(i)
And \((-10 + 3\mu-\lambda)3 + (–2μ -2 + 2λ)(-2) + (-3 - 2μ - 2λ)(-2) = 0,\)
i.e., \(17μ - 3λ = 20 \) ...(ii)
Solving (i) and (ii) for λ and μ, we get μ = 1, λ = -1.
The position vector of the points, at which they should be so that the distance between them is the shortest, is \(5\hat i + 4\hat j\) and \(-\hat i - 2\hat j - 3\hat k\)
\(\vec {PQ} = -6\hat i - 6 \hat j - 3 \hat k\)
The shortest distance = \(|\vec {PQ}|\)
\(= \sqrt{6^2 + 6^2 + 3^2}\)
\(= \sqrt 9\)
