Three Dimensional Geometry Question 114
Question: A square $ ABCD $ of diagonal 2a is folded along the diagonal $ AC $ so that the planes $ DAC $ and $ BAC $ are at right angle. The shortest distance between $ DC $ and $ AB $ is
[Kurukshetra CEE 1998]
Options:
A) $ \sqrt{2}a $
B) $ 2a/\sqrt{3} $
C) $ 2a/\sqrt{5} $
D) $(\sqrt{3}/2)a $
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Answer:
Correct Answer: B
Solution:
When folded coordinates will be $ D,(0,0,a);C,(a,,0,,0); $
$ A(-a,0,0);B,(0,-a,0) $
Equation $ DC\text{ is, }\frac{x}{a}=\frac{y}{b}=\frac{z-a}{-a} $
Equation $ AB\text{is,}\frac{x+a}{a}=\frac{y}{-a}=\frac{z}{b} $
$ \therefore $ Shortest distance $ =\frac{2a}{\sqrt{3}}, $ , (By formula).
BETA
