question_answer

If \[\theta \] is the acute angle between the diagonals of a cube, then which one of the following is correct?

A) \[\theta <30{}^\circ \]

B) \[\theta =60{}^\circ \]

C) \[30{}^\circ <\theta <60{}^\circ \]

D) \[\theta >60{}^\circ \]

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 \[(d{{r}_{3}})\] of these diagonals are: OE \[\left\langle (a-0),(a-0),(a-0) \right\rangle =(a,\,a,\,a)\]

\[FC\left\langle (-a,a,-a) \right\rangle ;GB\left\langle (-a,a,a) \right\rangle \] and

AD\[\left\langle (a,a,-a) \right\rangle \]

Their dcs are:

\[OE,\left\langle \frac{a}{\sqrt{{{a}^{2}}+{{a}^{2}}+{{a}^{2}}}},\frac{a}{\sqrt{{{a}^{2}}+{{a}^{2}}+{{a}^{2}}}},\frac{a}{\sqrt{{{a}^{2}}+{{a}^{2}}+{{a}^{2}}}} \right\rangle \]

\[=\left\langle \frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}} \right\rangle \]

\[AD,\left\langle \frac{a}{\sqrt{\Sigma {{a}^{2}}}},\frac{a}{\sqrt{\Sigma {{a}^{2}}}},\frac{-a}{\sqrt{\Sigma {{a}^{2}}}} \right\rangle \,=\,\left\langle \frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}} \right\rangle \]

Angle, \[\theta \], between AD and OE is given by

\[\cos \theta =\pm \frac{\frac{1}{\sqrt{3}}\times \frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}\times \frac{1}{\sqrt{3}}-\frac{1}{\sqrt{3}}\times \frac{1}{\sqrt{3}}}{\sqrt{\left\{ {{\left( \frac{1}{\sqrt{3}} \right)}^{2}}+\left( \frac{1}{{{\sqrt{3}}^{3}}} \right)+{{\left( \frac{1}{\sqrt{3}} \right)}^{2}} \right\}}\left\{ {{\left( \frac{1}{\sqrt{3}} \right)}^{2}}+{{\left( \frac{1}{\sqrt{3}} \right)}^{2}}+{{\left( \frac{1}{\sqrt{3}} \right)}^{2}} \right\}}\]

\[=\frac{\frac{1}{3}}{1\times 1}=\pm \frac{1}{3}\]

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\[\theta =90{}^\circ \]]