Three Dimensional Geometry Question 385
Question: If $ \theta $ is the acute angle between the diagonals of a cube, then which one of the following is correct?
Options:
A) $ \theta <30{}^\circ $
B) $ \theta =60{}^\circ $
C) $ 30{}^\circ <\theta <60{}^\circ $
D) $ \theta >60{}^\circ $
Show Answer
Answer:
Correct Answer: D
Solution:
[d] Let there be cube of side ?a? Co-ordinates of its vertices O, A, B, C, D, E, F have be marked in the figure. Diagonals are OE, FC, GB and AD. Direction ratios $ (dr_3) $ of these diagonals are: OE $ \langle (a-0),(a-0),(a-0) \rangle =(a,,a,,a) $ $ FC\langle (-a,a,-a) \rangle ;GB\langle (-a,a,a) \rangle $ and AD $ \langle (a,a,-a) \rangle $ Their dcs are: $ OE,\langle \frac{a}{\sqrt{a^{2}+a^{2}+a^{2}}},\frac{a}{\sqrt{a^{2}+a^{2}+a^{2}}},\frac{a}{\sqrt{a^{2}+a^{2}+a^{2}}} \rangle $ $ =\langle \frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}} \rangle $ $ AD,\langle \frac{a}{\sqrt{\Sigma a^{2}}},\frac{a}{\sqrt{\Sigma a^{2}}},\frac{-a}{\sqrt{\Sigma a^{2}}} \rangle ,=,\langle \frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}} \rangle $ Angle, $ \theta $ , between AD and OE is given by $ \cos \theta =\pm \frac{\frac{1}{\sqrt{3}}\times \frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}\times \frac{1}{\sqrt{3}}-\frac{1}{\sqrt{3}}\times \frac{1}{\sqrt{3}}}{\sqrt{{ {{( \frac{1}{\sqrt{3}} )}^{2}}+( \frac{1}{{{\sqrt{3}}^{3}}} )+{{( \frac{1}{\sqrt{3}} )}^{2}} }}{ {{( \frac{1}{\sqrt{3}} )}^{2}}+{{( \frac{1}{\sqrt{3}} )}^{2}}+{{( \frac{1}{\sqrt{3}} )}^{2}} }} $ $ =\frac{\frac{1}{3}}{1\times 1}=\pm \frac{1}{3} $ Since the cube is in positive octant, we take $ +\frac{1}{3}. $ So, $ \cos \theta =\frac{1}{3}\Rightarrow \theta >60’ $ [Since value of $ \cos \theta $ decreases as $ \theta $ increases in 0 to $ 90{}^\circ .\cos \theta =1 $ when $ \theta =0{}^\circ $ and $ \cos \theta =0 $ when $ \theta =90{}^\circ $ ]
BETA
