If a line makes angles $\alpha$, $\beta$, $\gamma$, $\delta$ with a cube's diagonals, then $\cos2\alpha+\cos2\beta+\cos2\gamma+\cos2\delta=-\frac43$

Question

If a line makes angles measuring $\alpha$, $\beta$, $\gamma$, $\delta$ with the four diagonals of a cube, prove that $$\cos2\alpha+\cos2\beta+\cos2\gamma+\cos2\delta=-\frac{4}{3} $$

My Approach:

• I found the direction ratios of all diagonals.
• Then Assumed the direction ratios of the line as l, m & n.
• Then I found all cosines...

D.R. are as follows,
$$\vec{OP}:(a,a,a),~~ \vec{AR}:(-a,a,a),~~ \vec{BS}:(a,-a,a), ~~ \vec{CQ}:(a,a,-a)$$

Diagram :

After that I don't know what to do

Answer 1

From the figure, the four diagonals are $~\vec{OP},~\vec{AR},~\vec{BS},~\vec{CQ}~$.

Direction ratios of $~\vec{OP}:a−o,a−o,a−o=a,a,a=1,1,1~$

Direction ratios of $~\vec{AR}:o−a,a−o,a−o=−a,a,a=−1,1,1~$

Direction ratios of $~\vec{BS}:a−o,o−a,a−o=a,−a,a=1,−1,1~$

Direction ratios of $~\vec{CQ}:a−o,a−o,o−a=a,a,−a=1,1,−1~$

$∴$ Direction cosine (DC)'s of $~\vec{OP}~$ are $~\dfrac{1}{\sqrt 3}~,\dfrac{1}{\sqrt 3},~\dfrac{1}{\sqrt 3}~$

DC's of $~\vec{AR}~$ are $~-\dfrac{1}{\sqrt 3}~,\dfrac{1}{\sqrt 3},~\dfrac{1}{\sqrt 3}~$

DC's of $~\vec{BS}~$ are $~\dfrac{1}{\sqrt 3}~,-\dfrac{1}{\sqrt 3},~\dfrac{1}{\sqrt 3}~$

DC's of $~\vec{CQ}~$ are $~\dfrac{1}{\sqrt 3}~,-\dfrac{1}{\sqrt 3},~-\dfrac{1}{\sqrt 3}~$

Let $~l,~m,~n~$ be DC's of line $($and hence $~l^2+m^2+n^2=1~)$ and line makes angle $~α~$ with $~\vec{OP}~$. So $$\cos α=l\left(\dfrac{1}{\sqrt 3}\right)+m\left(\dfrac{1}{\sqrt 3}\right)+n\left(\dfrac{1}{\sqrt 3}\right)=\dfrac{l+m+n}{\sqrt 3}$$

Similarly $$\cos \beta=l\left(-\dfrac{1}{\sqrt 3}\right)+m\left(\dfrac{1}{\sqrt 3}\right)+n\left(\dfrac{1}{\sqrt 3}\right)=\dfrac{-l+m+n}{\sqrt 3}~,$$ ​ $$\cos \gamma=l\left(\dfrac{1}{\sqrt 3}\right)+m\left(-\dfrac{1}{\sqrt 3}\right)+n\left(\dfrac{1}{\sqrt 3}\right)=\dfrac{l-m+n}{\sqrt 3}~,$$

$$\cos \delta=l\left(\dfrac{1}{\sqrt 3}\right)+m\left(\dfrac{1}{\sqrt 3}\right)+n\left(-\dfrac{1}{\sqrt 3}\right)=\dfrac{l+m-n}{\sqrt 3}$$

Now

\begin{equation} \cos2\alpha+\cos2\beta+\cos2\gamma+\cos2\delta\\ =2\left(\cos^2\alpha+\cos^2\beta+\cos^2\gamma+\cos^2\delta\right)-4\\ =\dfrac 23\left[(l+m+n)^2+(-l+m+n)^2+(l-m+n)^2+(l+m-n)^2\right]-4\\ =\dfrac 23\left[4(l^2+m^2+n^2)\right]-4\\ =4\left[\dfrac 23-1\right]\\ =-\dfrac 43 \end{equation}