Question

The shortest distance between the lines
\(\frac{x-4}{4}=\frac{y+2}{5}=\frac{z+3}{3}\)
and
\(\frac{x-1}{3}=\frac{y-3}{4}=\frac{z-4}{2}\)
is :

• A. \(3\sqrt{6}\)
• B. \(6\sqrt{2}\)
• C. \(6\sqrt{3}\)
• D. \(2\sqrt{6}\)

Hint:

Use \(d=\dfrac{|(\vec{b}-\vec{a})\cdot(\vec{d_1}\times\vec{d_2})|}{|\vec{d_1}\times\vec{d_2}|}\) with the direction vectors \((4,5,3)\) and \((3,4,2)\).

Answer 1

The Correct Option is A

Take \(P(4+4s,\,-2+5s,\,-3+3s)\) on the first line and \(Q(1+3t,\,3+4t,\,4+2t)\) on the second. The shortest distance is the length of the common perpendicular \(PQ\).
\(\overrightarrow{PQ}=(-3+3t-4s,\;5+4t-5s,\;7+2t-3s)\).
\(\overrightarrow{PQ}\cdot(4,5,3)=0\Rightarrow 34+38t-50s=0\Rightarrow 19t-25s=-17\).
\(\overrightarrow{PQ}\cdot(3,4,2)=0\Rightarrow 25+29t-38s=0\Rightarrow 29t-38s=-25\).
Multiplying the first by 29 and the second by 19 and subtracting: \(-3s=-18\Rightarrow s=6\), then \(19t=-17+150=133\Rightarrow t=7\).
\(\overrightarrow{PQ}=(-3+21-24,\;5+28-30,\;7+14-18)=(-6,3,3)\).
\(|PQ|=\sqrt{36+9+9}=\sqrt{54}=3\sqrt{6}\).
Answer: (1) \(3\sqrt{6}\)