Diagonals of a cube

A cube is rectangular parallelepiped 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 cosines of the diagonal OE which is the like joining two points O and E are

$\frac{1}{√3}$ , $\frac{1}{√3}$ , $\frac{1}{√3}$

Similarly, the direction cosines of AF, BG and CD are $\frac{-1}{√3}$ , $\frac{1}{√3}$ , $\frac{1}{√3}$ ; $\frac{1}{√3}$ , $\frac{-1}{√3}$ , $\frac{1}{√3}$ ; $\frac{1}{√3}$ , $\frac{1}{√3}$ , $\frac{-1}{√3}$ respectively.

Let $l, m, n$ be the direction cosines of the given line which makes makes angles α, β, γ, 𝛿 with OE, AF, BG, CD respectively.

Then, cos α = $\frac{1}{√3}$ ( $l + m + n$ ); cos β = $\frac{1}{√3}$ ( $-l + m + n$ )

cos γ = $\frac{1}{√3}$ ( $l - m + n$ ); cos 𝛿 = $\frac{1}{√3}$ ( $l + m - n$ )

Squaring and adding, we get

$cos^2 α$ + $cos^2 β$ + $cos^2 γ$ + \(cos^2 𝛿\) = $\frac{1}{3}$ [ $(l + m + n)^2$ + $(-l + m + n)^2$ + $(l - m + n)^2$ + $(l + m - n)^2$ ]

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

As $l^2 + m^2 + n^2$ = 1

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

Where $a$ = 4, $b$ = 3

$a + b$ = $4 + 3$ = $\boxed{7}$ .

Therefore the Answer is $\boxed{7}$ .

Consider the given line to be one of the four diagonals.
If side of the cube is $a$ , lengths of half diagonals is $\frac{a\sqrt{3}}{2}$
By using cosine rule --
We get $cos\alpha=cos\beta=cos\gamma=\frac{1}{3}\text{ \& }cos\delta=1$
so, $\frac{a}{b}=\frac{4}{3}\text{ hence, }\boxed{a+b=7}$