Diagonals of a Cube

Let the unit vectors of four diagnols be :

$\hat { { r }_{ 1 } } =\quad \frac { 1 }{ \sqrt { 3 } } (\hat { i } +\hat { j } +\hat { k } )$

$\hat { { r }_{ 2 } } =\quad \frac { 1 }{ \sqrt { 3 } } (\hat { i } -\hat { j } +\hat { k } )$

$\hat { { r }_{ 3 } } =\quad \frac { 1 }{ \sqrt { 3 } } (-\hat { i } +\hat { j } +\hat { k } )$

$\hat { { r }_{ 4 } } =\quad \frac { 1 }{ \sqrt { 3 } } (-\hat { i } -\hat { j } +\hat { k } )$

Any arbitary unit vector can be expressed as :

$\hat { { a } } =\quad l\hat { i } +m\hat { j } +n\hat { k }$

$l,m,n$ are directional cosines of $a$

Now $\hat { { a } } .\hat { { r }_{ 1 } } = cos(\alpha)$

$\hat { { a } } .\hat { { r }_{ 1 } } = cos(\beta)$

$\hat { { a } } .\hat { { r }_{ 1 } } = cos(\gamma)$

$\hat { { a } } .\hat { { r }_{ 1 } } = cos(\delta)$

Our desired thing becomes :

$({ \hat { { a } } .\hat { { r }_{ 1 } } ) }^{ 2 }+{ ({ \hat { { a } } .\hat { { r }_{ 2 } } ) } }^{ 2 }+{ ({ \hat { { a } } .\hat { { r }_{ 3 } } ) } }^{ 2 }{ +({ \hat { { a } } .\hat { { r }_{ 4 } } ) } }^{ 2 }$

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

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

Also we know that ${ l }^{ 2 }+{ m }^{ 2 }+{ n }^{ 2 }=1$

$S=\frac{4}{3}$

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

If there is a line that makes angles with the 4 diagonals, it has to intercept the intersection of the diagonals, leaving the same angle to all of them. That leaves us with $4 \cos^2x$ .

The diagonal of the square can be expressed by the $\sqrt 3 * \text{ side}$ . And $\cos x=\dfrac{\text{ side}}{\sqrt 3 * \text{ side}} \Rightarrow 4 \cos^2 x=\dfrac 43$ .

Hence the required value is $4+3=7. \square$

Okay there's a whole method to find this value but here let us just be smarter as far as the value is concerned. We can think of a very special case when $\alpha$ , $\beta$ , $\gamma$ & $\delta$ are all equal & equal to $\theta$ (say). In this case, our line will start from O & will meet the middle point or the meeting point of the diagonals of the square $P_1P_2P_3P_4$ . Let this point be G.

Now let $2a$ be the length of the sides of cube. Using Pythagoras theorem in $\triangle P_1GP_4$ we find $P_1G=√2a$ . Now since $GO=a$ (why). From right triangle $P_1GO$ $\tan\theta=\frac{P_1G}{GO}=√2$ $\cos^2\theta=\frac{1}{\sec^2\theta}=\frac{1}{1+\tan^2\theta}=\frac{1}{3}$ Hence our required value reduces to $1/3+1/3+1/3+1/3=4/3$ $\Rightarrow 4+3=\boxed7$

Let $A, B, C, D, G$ be the vertices of the base of the cube, and $(O, x, y, z,)\,$ be a system of cartesian coordinates in $\mathbb{{R}^3}$ with $A \equiv O,\, x \parallel {(B - O)},\, y \parallel {(D - O)},\, z \perp {(x, y)},$ $\,G$ the point of intersection of the four diagonals, and $L$ the lenght of the side cof the cube ; we have:

$O(0, 0, 0)$

$B(L, 0, 0)$

$C(L, L, 0)$

$D(0, L, 0)$

$G(\frac L 2, \frac L 2, \frac L 2),\,$ so:

$|GO| = |GB| = |GC| = |GD| = {\frac {\sqrt 3} 2}L,\,$ and the unit versors of the diagonals are: $\vec {u_i} = \frac {G - {A_i}} {|G{C_i}|}\,$

$\vec {u_A} = {\frac 1 {\sqrt 3}}(1, 1, 1)$

$\vec {u_B} = {\frac 1 {\sqrt 3}}(-1, 1, 1)$

$\vec {u_C} = {\frac 1 {\sqrt 3}}(1, -1, 1)$

$\vec {u_D} = {\frac 1 {\sqrt 3}}(-1, -1, 1)$

if $\vec{u_l} = (cos{{\theta}_ x},\,cos{{\theta}_ y},\,cos{{\theta}_ z})$ is the unit vector of a generic stright line, we have:

${\vec{u_l}}\cdot{\vec{u_l}} = |\vec {u_l}|^2=1$

$cos(\alpha) = {\vec {u_A}}\cdot{\vec{u_l}}$

$cos(\beta) = {\vec {u_B}}\cdot{\vec{u_l}}$

$cos(\gamma) = {\vec {u_C}}\cdot{\vec{u_l}}$

$cos(\delta) = {\vec {u_D}}\cdot{\vec{u_l}}$

substituting, squaring, summing and simplyfing we obtain:

${cos^2}(\alpha)+{cos^2}(\beta)+{cos^2}(\gamma)+{cos^2}(\delta)={\frac 4 3}({cos^2}({\theta}_ x)+{cos^2}({\theta}_ y)+{cos^2}({\theta}_ z)) =$

$\boxed {\frac 4 3}$