Question

If a line makes angles $\alpha ,\beta ,\gamma$ and $\delta$ with the diagonals of a cube, Then, ${\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma +{\cos }^2\delta =\frac{a}{b}$, where $a$ and $b$ are in lowest form, find $a+b$

• A. $7$
• B. $6$
• C. $8$
• D. None of these

Answer 1

The correct option is A $7$

A cube is a rectangular parallelopiped having equal length, breadth and height. Let OADBFEGC be the cube with each side of length a units.

The four diagonals are OE, AF, BG and CD.

The direction \cos ines of the diagonal OE which is the line joining two points O and E are

$\frac{a-0}{\sqrt{a^2+a^2+a^2}},\frac{a-0}{\sqrt{a^2+a^2+a^2}},\frac{a-0}{\sqrt{a^2+a^2+a^2}}$

i.e. $\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}$

Similarly, the direction \cos ines of AF, BG and CD are $(-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}});(\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}});(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}})$, respectively.

Let l, m, n be the direction \cos ines of the given line which makes angles $\alpha ,\beta ,\gamma ,\delta$ with OE, AF, BG, CD, respectively. Then

$\cos \alpha =\frac{1}{\sqrt{3}}(l+m+n);\cos \beta =\frac{1}{\sqrt{3}}(-l+m+n)$;

$\cos \gamma =\frac{1}{\sqrt{3}}(l-m+n);\cos \delta =\frac{1}{\sqrt{3}}(l+m+n)$;

Squaring and adding, we get

$cs{o}^2\alpha +{\cos }^2\beta +{\cos }^2\gamma +{\cos }^2\delta =\frac{1}{3}\left[\right(1+m+n{)}^2+(-1+m+n{)}^2+(l-m+n{)}^2+(l+m+n{)}^2]=\frac{1}{3}\left[4\right(l^2+m^2+n^2\left)\right]=\frac{4}{3}$

(as $l^2+m^2+n^2=1)$