Question

If a line makes angles α, β, γ, δ with four diagonals of a cube, then cos² α + cos² β + cos² γ + cos² δ is equal to

(a) $\frac{1}{3}$

(b) $\frac{2}{3}$

(c) $\frac{4}{3}$

(d) $\frac{8}{3}$

Answer 1

$\left(\mathrm{c}\right)\frac{4}{3}$




$$\begin{gathered} \mathrm{Let}a\mathit{}\mathrm{be}\mathrm{the}\mathrm{length}\mathrm{of}\mathrm{an}\mathrm{edge}\mathrm{of}\mathrm{the}\mathrm{cube}\mathrm{and}\mathrm{let}\mathrm{one}\mathrm{corner}\mathrm{be}\mathrm{at}\mathrm{the}\mathrm{origin}\mathrm{as}\mathrm{shown}\mathrm{in}\mathrm{the}\mathrm{figure}.\mathrm{Clearly},OP\mathit{,}\mathit{}AR\mathit{,}\mathit{}BS\mathit{}\mathrm{and}CQ\mathit{}\mathrm{are}\mathrm{the}\mathrm{diagonals}\mathrm{of}\mathrm{the}\mathrm{cube}. \\ \mathrm{The}\mathrm{direction}\mathrm{ratios}\mathrm{of}OP\mathit{,}\mathit{}AR\mathit{,}\mathit{}BS\mathit{}\mathrm{and}CQ\mathit{}\mathrm{are} \\ a-0,a\mathit{-}0,a-0,\mathrm{i}.\mathrm{e}.a\mathit{,}\mathit{}a\mathit{,}\mathit{}a \\ 0-a,a-0,a-0,\mathrm{i}.\mathrm{e}.-a\mathit{,}\mathit{}a\mathit{,}\mathit{}a \\ a-0,0-a,a-0,\mathrm{i}.\mathrm{e}.a\mathit{,}\mathit{}\mathit{-}a\mathit{,}\mathit{}a \\ a-0,a-0,0-a,\mathit{}\mathrm{i}.\mathrm{e}.a\mathit{,}\mathit{}a\mathit{,}\mathit{-}a\mathit{} \\ \mathrm{Let}\mathrm{the}\mathrm{direction}\mathrm{ratios}\mathrm{of}\mathrm{a}\mathrm{line}\mathrm{be}\mathrm{proportional}\mathrm{to}l\mathit{,}\mathit{}m\mathrm{and}n.\mathrm{Suppose}\mathrm{this}\mathrm{line}\mathrm{makes}\mathrm{angles}\alpha ,\mathrm{\beta },\mathrm{\gamma }\mathrm{and}\mathrm{\delta }\mathrm{with}OP\mathit{,}\mathit{}AR\mathit{,}\mathit{}BS\mathit{}\mathrm{and}CQ,\mathit{}\mathrm{respectively}. \\ \mathrm{Now},\mathrm{\alpha }\mathrm{is}\mathrm{the}\mathrm{angle}\mathrm{between}OP\mathit{}\mathrm{and}\mathrm{the}\mathrm{line}\mathrm{whose}\mathrm{direction}\mathrm{ratios}\mathrm{are}\mathrm{proportional}\mathrm{to}l\mathit{,}\mathit{}m\mathrm{and}\mathit{}n\mathit{.} \\ \cos \mathrm{\alpha }=\frac{a.l+a.m+a.n}{\sqrt{a^2+a^2+a^2}\sqrt{l^2+m^2+n^2}}\Rightarrow \cos \mathrm{\alpha }=\frac{l\mathit{}+m\mathit{}+n}{\sqrt{3}\sqrt{{\mathrm{l}}^2+{\mathrm{m}}^2+{\mathrm{n}}^2}} \\ \mathrm{Since}\mathrm{\beta }\mathrm{is}\mathrm{the}\mathrm{angle}\mathrm{between}AR\mathit{}\mathrm{and}\mathrm{the}\mathrm{line}\mathrm{with}\mathrm{direction}\mathrm{ratios}\mathrm{proportional}\mathrm{to}\mathit{}l\mathit{,}\mathit{}m\mathrm{and}\mathit{}n,\mathrm{we}\mathrm{get} \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\cos \mathit{}\mathrm{\beta }=\frac{-a\mathit{.}l+a\mathit{.}m+a\mathit{.}n}{\sqrt{a^2+a^2+a^2}\sqrt{l^2+m^2+n^2}}\Rightarrow \cos \mathit{}\mathrm{\beta }=\frac{-l+m+n}{\sqrt{3}\sqrt{l^2+m^2+n^2}} \\ \mathrm{Similarly}, \\ \cos \mathrm{\gamma }=\frac{a\mathit{.}l\mathit{-}a\mathit{.}m+a\mathit{.}n}{\sqrt{a^2+a^2+a^2}\sqrt{l^2+m^2+n^2}}\Rightarrow \cos \mathrm{\gamma }=\frac{l\mathit{}-\mathit{}m\mathit{}+\mathit{}n}{\sqrt{3}\sqrt{l^2+m^2+n^2}} \\ \cos \mathrm{\delta }=\frac{a\mathit{.}l+a\mathit{.}m-a\mathit{.}n}{\sqrt{a^{\mathit{2}}\mathit{+}{a}^{\mathit{2}}\mathit{+}{a}^{\mathit{2}}}\sqrt{l^{\mathit{2}}\mathit{+}{m}^{\mathit{2}}\mathit{+}{n}^{\mathit{2}}}}\Rightarrow \cos \mathrm{\delta }=\frac{l+m\mathit{}-\mathit{}n}{\sqrt{3}\sqrt{l^2+m^2+n^2}} \\ {\cos }^2\mathrm{\alpha }+{\cos }^2\mathrm{\beta }+{\cos }^2\mathrm{\gamma }+{\cos }^2\mathrm{\delta } \\ =\frac{{\left(l\mathit{+}m\mathit{+}n\right)}^2}{3\left(l^2+m^2+n^2\right)}+\frac{{\left(-l+m+n\right)}^2}{3\left(l^2+m^2+n^2\right)}+\frac{{\left(I-m+n\right)}^2}{3\left(l^2+m^2+n^2\right)}+\frac{{\left(l+m\mathit{}-\mathit{}n\right)}^2}{\sqrt{3}\sqrt{l^2+m^2+n^2}} \\ =\frac{1}{3\left(l^2+m^2+n^2\right)}\left\{{\left(l+m+n\right)}^2+{\left(-l+m+n\right)}^2+{\left(I-m+n\right)}^2+{\left(l+m\mathit{}-n\right)}^2\right\} \\ =\frac{1}{3\left(l^2+m^2+n^2\right)}4\left(l^2+m^2+n^{\mathit{2}}\right)=\frac{4}{3} \end{gathered}$$