Three Dimensional Geometry Question 45
Question: The shortest distance between the lines $ \frac{x-3}{3}=\frac{y-8}{-1}=\frac{z-3}{1} $ and $ \frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-6}{4} $ is
[RPET 2001; MP PET 2002]
Options:
A) $ \sqrt{30} $
B) $ 2\sqrt{30} $
C) $ 5\sqrt{30} $
D) $ 3\sqrt{30} $
Show Answer
Answer:
Correct Answer: D
Solution:
S.D. $ = $ $ \frac{ \begin{vmatrix} 6 & 15 & -3 \\ 3 & -1 & 1 \\ -3 & 2 & 4 \\ \end{vmatrix} }{\sqrt{{{(-4-2)}^{2}}+{{(12+3)}^{2}}+{{(6-3)}^{2}}}} $ $ =\frac{270}{\sqrt{270}}=\sqrt{270} $ = $ 3\sqrt{30} $ .
BETA
