Let P(α,β,γ) be the point on the line 2x−1=−3y+1=1z at a distance 414 from the point (1,−1,0) and nearer to the origin. Then the shortest distance between the lines 1x−α=2y−β=3z−γ and 2x+5=1y−10=1z−3 is equal to:
Solutions · 1
Official solution
Point on the line: (1+2t,−1−3t,t). Distance2 from (1,−1,0): 4t2+9t2+t2=14t2=16⋅14⇒t=±4.
Candidates: (9,−13,4) and (−7,11,−4). Distances2 from origin: 266 and 186 respectively, so the nearer point is
\[(\alpha,\beta,\gamma)=(-7,11,-4).\]
Now find the shortest distance between a=(−7,11,−4) (direction b=(1,2,3)) and c=(−5,10,3) (direction d=(2,1,1)):
\[\vec b\times\vec d=(1,2,3)\times(2,1,1)=(2\cdot1-3\cdot1,\ 3\cdot2-1\cdot1,\ 1\cdot1-2\cdot2)=(-1,5,-3).\]
\[\vec c-\vec a=(2,-1,7).\]
\[\text{S.D.}=\frac{|(\vec c-\vec a)\cdot(\vec b\times\vec d)|}{|\vec b\times\vec d|}=\frac{|-2-5-21|}{\sqrt{1+25+9}}=\frac{28}{\sqrt{35}}=\frac{4\sqrt{35}}{5}=4\sqrt{\frac75}.\]