Answer:
Step-by-step explanation:
Let the I line be PQ and II line be RS
Direction ratios of PQ = (9,0,0) and direction ratios of RS = (0, -4, -12)
Distance between skew lines
=(a1-a2).(b1xb2)/|b1xb2|
where a1 and a2 are points
Here a1-a2 = (-3,2,-4) taking P and Q as points.
b1 = (9,0,0) and b2 = (0,-4,-12)
Shortest Distance = (a2-a1).(b1xb2)/|b1xb2|
De![\left[\begin{array}{ccc}-3&2&-4\\9&0&0\\0&-4&-12\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-3%262%26-4%5C%5C9%260%260%5C%5C0%26-4%26-12%5Cend%7Barray%7D%5Cright%5D)
Numerator = -9(-24-16)=360
Denominator vector = i(0)-j(-108)+k(-36)
Modulus of vector =\sqrt{108^2+36^2} =36\sqrt{10}
Hence Shortest distance = \sqrt{10}