Question

If a line makes $\alpha ,\beta ,\gamma \text{and}\delta$ from the diagonals of a cube then ${\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma +{\cos }^2\delta =\frac{a}{3}$. Find $a$

Answer 1

$LetOA=OB=OC=a$

Then the co-ordinates of $O,A,B\&C$ are

$(0,0,0),(a,0,0),(0,a,0),(0,0,a).$ and that of $P,L,MandN$ are$(a,a,a),(0,a,a),(0,0,a),(a,a,0)$ respectively.

The four diagonals are $UP,AL,BM,CN.$

Direction cosines of $OP$ are proportional to$a-0,a-0,a-0i.e,(a,0,a),i.e.,1,1,1.$

Similarily,

Direction cosines of $AL$ are proportional to $0-a,a-0,a0i.e.,-1,1,1.$

Direction cosines of $BM$ are proportional to $1,1,-1$

$\therefore$ Direction cosines of $OP$ are $\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}$

Direction cosines of $AL$ are $\frac{1}{\sqrt{3}},\frac{-1}{\sqrt{3}},\frac{1}{\sqrt{3}}$

Direction cosines of $BM$ are $\frac{1}{\sqrt{3}},\frac{-1}{\sqrt{3}},\frac{1}{\sqrt{3}}$

Direction cosines of $CN$ are $\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{-1}{\sqrt{3}}$

Let l,m,n be direction cosine of the line.

$\therefore$ The line makes angle $\alpha$ with $OP.$

$\cos \alpha =l\left(\frac{1}{\sqrt{3}}\right)+m\left(\frac{1}{\sqrt{3}}\right)+n\left(\frac{1}{\sqrt{3}}\right)$

$⟹\cos \alpha =\frac{l+m+n}{\sqrt{3}}-\left(i\right)$

Similarily$,\cos \beta =\frac{-l+m+n}{\sqrt{3}}-\left(ii\right)$

and$\cos \gamma =\frac{l-m+n}{\sqrt{3}}-\left(iii\right)$

$\cos \delta =\frac{l+m-n}{\sqrt{3}}-\left(iv\right)$

Squaring and adding $\left(i\right),\left(ii\right),\left(iii\right)\&\left(iv\right)$

$⟹\cos {}^2\alpha +\cos {}^2\beta +\cos {}^2\gamma +\cos {}^2\delta$

$=\frac{1}{3}[{(l+m+n)}^2+{(-l+m+n)}^2+{(l-m+n)}^2+{(l+m-n)}^2]$

$=\frac{1}{3}[4l^2+4m^2+4n^2]=\frac{4}{3}[l^2+m^2+n^2]$

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

$\therefore a=4$