Question

A line makes angles $\alpha ,\beta ,\gamma ,\delta$ with four diagonals of a cube. then $\sum _{rϵ\{\alpha ,\beta ,\gamma ,\delta \}}co{s}^2\left(r\right)=$

• A. $\frac{-4}{3}$
• B. $\frac{4}{3}$
• C. $\frac{3}{4}$
• D. $\frac{3}{5}$

Answer 1

Answer: (A), (B)

The correct options are
A $\frac{-4}{3}$
B $\frac{4}{3}$

Let $a$ be the length of an edge of the cube and one of the end is at origin.

Clearly, $OP,AR,BS$ and $CQ$ are the diagonals of the cube.

The direction ratios of $OP,AR,BS$ and $CQ$ are:

$a-0,a-0,a-0i.e.a,a,a$

$0-a,a-0,a-0i.e.-a,a,a$

$a-0,0-a,a-0i.e.a,-a,a$

and $a-0,a-0,0-ai.e,a,a,-a$

Let the direction ratios of a line be proportional to $l,m,n$

Suppose this line makes angle $\alpha ,\beta ,\gamma$ and $\delta$ with $OP,AR,BS$ and $CQ$ respectively.

Now $\alpha$ is the angle between $OP$ and the line whose direction ratios are proportional to $l,m,n$

$\therefore \cos \alpha =\frac{a\cdot l+a\cdot m+a\cdot n}{\sqrt{a^2+a^2+a^2}\sqrt{l^2+m^2+n^2}}$

$⟹\cos \alpha =\frac{l+m+n}{\sqrt{3}\sqrt{l^2+m^2+n^2}}$ ....... (1)

Now $\beta$ is the angle between $AR$ and the line whose directional ratios are proportional to $l,m,n$

$\therefore \cos \beta =\frac{-a\cdot l+a\cdot m+a\cdot n}{\sqrt{(-a{)}^2+a^2+a^2}\sqrt{l^2+m^2+n^2}}$

$⟹$$\cos \beta =\frac{-l+m+n}{\sqrt{3}\sqrt{l^2+m^2+n^2}}$ ..... (2)

Similarly,

$\cos \gamma =\frac{l-m+n}{\sqrt{3}\sqrt{l^2+m^2+n^2}}$ ........... (3)

and $\cos \delta =\frac{l+m-n}{\sqrt{3}\sqrt{l^2+m^2+n^2}}$ ....... (4)

$\therefore$ from (1),(2),(3) and (4) we have

$\sum _{rϵ\{\alpha ,\beta ,\gamma ,\delta \}}co{s}^2\left(r\right)$

$={\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma +{\cos }^2\delta$

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

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

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