question_answer

If a line makes angles \[\alpha ,\beta ,\gamma ,\delta \] with four diagonals of a cube, then the value of \[{{\sin }^{2}}\alpha +{{\sin }^{2}}\beta +\] \[{{\sin }^{2}}\gamma +{{\sin }^{2}}\delta \] is                                [MP PET 2004]

A)            \[\frac{4}{3}\]

B)            1

C)            \[\frac{8}{3}\]

D)            \[\frac{7}{3}\]

Correct Answer: C

Solution :

           Let side of the cube = a          Then OG, BE and AD, CF will be four diagonals.          d.r.?s of OG = a, a, a = 1, 1, 1          d.r.?s of BE = ?a, ?a, a = 1, 1, ?1          d.r.?s of AD = ?a, a, a = ?1, 1, 1          d.r.?s of CF = a, ?a, a = 1, ?1, 1          Let d.r.?s of line be l, m, n.          Therefore angle between line and diagonal          \[\cos \alpha =\frac{l+m+n}{\sqrt{3}},\,\cos \beta =\frac{l+m-n}{\sqrt{3}},\,\]          \[\cos \gamma =\frac{-l+m+n}{\sqrt{3}},\,\cos \delta =\frac{l-m+n}{\sqrt{3}}\]          Þ  \[{{\cos }^{2}}\alpha +{{\cos }^{2}}\beta +{{\cos }^{2}}\gamma +{{\cos }^{2}}\delta \]               \[=\frac{1}{3}[{{(l+m+n)}^{2}}+{{(l+m-n)}^{2}}+{{(-l+m+n)}^{2}}+{{(l-m+n)}^{2}}]\]            \[=\frac{4}{3}\]Þ \[{{\sin }^{2}}\alpha +{{\sin }^{2}}\beta +{{\sin }^{2}}\gamma +{{\sin }^{2}}\delta =\frac{8}{3}\].