Question

If a line makes $\alpha ,\beta ,\gamma ,\delta$ angles with four diagonals of a cube, then the value ${\sin }^2\alpha +{\sin }^2\beta +{\sin }^2\gamma +{\sin }^2\delta$ is

• A. $\frac{4}{3}$
• B. $1$
• C. $\frac{8}{3}$
• D. $\frac{7}{3}$

Answer 1

The correct option is C

$\frac{8}{3}$

Explanation for the correct option:

Step 1. Draw the figure of cube according to the question:

From the figure, we can see

Four diagonals of a cube are $OP,AM,CL,BN$

Now,

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

Direction cosines of $AM=-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}$

Direction cosines of $CL=\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}$

Direct cosines of $BN=\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}.\frac{1}{\sqrt{3}}$

Step 2. Let $l,m,n$ be the direction cosines of the line which makes angle $\alpha ,\beta ,\gamma ,\delta$, we get

$\cos \alpha =\frac{l+m+n}{\sqrt{3}}$

$\cos \beta =\frac{-l+m+n}{\sqrt{3}}$

$\cos \gamma =\frac{l+m-n}{\sqrt{3}}$

$\cos \delta =\frac{l-m+n}{\sqrt{3}}$

Now,

$\begin{array}{rcl}{\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma +{\cos }^2\delta & =& \frac{1}{3}\left[{\left(l+m+n\right)}^2+{\left(-l+m+n\right)}^2+{\left(l+m+n\right)}^2+{\left(l-m+n\right)}^2\right]\\ & =& \frac{1}{3}\left[4\left(l^2+m^2+n^2\right)\right]\\ & =& \mathbf{}\frac{4}{3}\end{array}$$\begin{array}{rcl}\therefore {\sin }^2\alpha +{\sin }^2\beta +{\sin }^2\gamma +{\sin }^2\delta & =& \left(1-{\cos }^2\alpha \right)+\left(1-{\cos }^2\beta \right)+\left(1-{\cos }^2\gamma \right)+\left(1-{\cos }^2\delta \right)\\ & =& 4-({\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma +{\cos }^2\delta )\\ & =& 4-\frac{4}{3}\\ & =& \frac{\mathbf{8}}{\mathbf{3}}\end{array}$

Hence, option ‘C’ is Correct.