Question

If the shortest distance between the lines \(3x+2y-4z-5=0=5x-7y-17z+2\) and \(\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+1}{2}\) is \(\frac{10}{\sqrt{k}}\), then \(k\) is equal to ________.

Hint:

Find the direction and a point of the first line from the two planes, then use \(\frac{|(\vec a_2-\vec a_1)\cdot(\vec d_1\times\vec d_2)|}{|\vec d_1\times\vec d_2|}\).

Answer 1

Correct Answer: 59

Find the plane that contains line 1 and is parallel to line 2; the required distance is the distance of any point of line 2 from this plane.
Line 1 has direction \(\vec d_1=(2,-1,1)\) (cross product of the two plane normals) and passes through \((1,1,0)\). Line 2 has direction \((3,-5,2)\).
The normal of the required plane is \(\vec d_1\times(3,-5,2)=(3,-1,-7)\). The plane through \((1,1,0)\): \(3(x-1)-(y-1)-7z=0\), i.e. \(3x-y-7z-2=0\).
Distance of the point \((2,1,-1)\) of line 2 from it: \(\frac{|6-1+7-2|}{\sqrt{9+1+49}}=\frac{10}{\sqrt{59}}\).
Comparing with \(\frac{10}{\sqrt k}\) gives \(k=59\).
Answer: 59