Three-Dimensional-Geometry Question 438
Question: 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]
Options:
A) $ \frac{4}{3} $
B) 1
C) $ \frac{8}{3} $
D) $ \frac{7}{3} $
Show Answer
Answer:
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} $ .
BETA
