Question

A line makes angle ${\theta }_1,{\theta }_2,{\theta }_3,{\theta }_4$ with the diagonals of the cube. Show that ${\cos }^2{\theta }_1+{\cos }^2{\theta }_2+{\cos }^2{\theta }_3+{\cos }^2{\theta }_4=\frac{4}{3}$?

Answer 1

Let side of cube be 'a'

Dr's of a diagonal OG = a ,a ,a

Dr's of a diagonal CD = a ,a , - a

Dr's of a diagonal BE = a , - a ,a

Dr's of a diagonal AF = - a ,a ,a

$nowD{c}^{\prime }sofdiagonalOG=\left(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right)$

$nowD{c}^{\prime }sofdiagonalCD=\left(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{-1}{\sqrt{3}}\right)$

$nowD{c}^{\prime }sofdiagonalBE=\left(\frac{1}{\sqrt{3}},\frac{-1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right)$

$nowD{c}^{\prime }sofdiagonalAF=\left(\frac{-1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right)$

lets dcs of given line which is making

${\theta }_1,{\theta }_2,{\theta }_3,{\theta }_4,$ with the diagonal be l, m, n now applying between two angle

so

$$\begin{array}{c}cos{\theta }_1=l\left(\frac{1}{\sqrt{3}}\right)+m\left(\frac{1}{\sqrt{3}}\right)+n\left(\frac{1}{\sqrt{3}}\right)\to \left(1\right)\\ cos{\theta }_2=l\left(\frac{1}{\sqrt{3}}\right)-m\left(\frac{1}{\sqrt{3}}\right)+n\left(\frac{1}{\sqrt{3}}\right)\to \left(2\right)\\ cos{\theta }_3=l\left(\frac{1}{\sqrt{3}}\right)+m\left(\frac{1}{\sqrt{3}}\right)-n\left(\frac{1}{\sqrt{3}}\right)\to \left(3\right)\\ cos{\theta }_4=-l\left(\frac{1}{\sqrt{3}}\right)+m\left(\frac{1}{\sqrt{3}}\right)+n\left(\frac{1}{\sqrt{3}}\right)\to \left(4\right)\end{array}$$

squaring and adding equation (1), (2) , (3) (4)

$co{s}^2{\theta }_1+co{s}^2{\theta }_2+co{s}^2{\theta }_3+co{s}^2{\theta }_4=$

$\frac{1}{3}\left[l^2+m^2+n^2+2lm+2mn+2\ln +l^2+m^2+n^2+2lm-2mn-2\ln +l^2+m^2+n^2-2lm-2mn+2\ln +l^2+m^2+n^2-2lm+2mn-2\ln \right]$

$\frac{1}{3}\left[4\left(l^2+m^2+n^2\right)\right]$

$\frac{1}{3}\left(4\right)\to \left(l^2+m^2+n^2\right)=1$

$\frac{4}{3}$

$co{s}^2{\theta }_1+co{s}^2{\theta }_2+co{s}^2{\theta }_3+co{s}^2{\theta }_4=\frac{4}{3}$