Question

A line makes angles $a,b,c,d$ with the four diagonals of a cube, then ${\cos }^{2}a+{\cos }^{2}b+{\cos }^{2}c+{\cos }^{2}d=$

• A. $\frac{1}{3}$
• B. $\frac{4}{3}$
• C. $\frac{2}{3}$
• D. $\frac{5}{3}$

Answer 1

The correct option is B $\frac{4}{3}$

Four diagonal of the cube are $AL,BM,CN\text{and}OP$
(putting $a=b=c$)

Direction cosine of the four diagonals are
$OP(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}),AL(\frac{-1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}})BM(\frac{1}{\sqrt{3}},\frac{-1}{\sqrt{3}},\frac{1}{\sqrt{3}}),CN(\frac{-1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{-1}{\sqrt{3}})$
Let the D.C.'s of the line is $l,m,n$ then
$\cos a=\frac{l}{\sqrt{3}}+\frac{m}{\sqrt{3}}+\frac{n}{\sqrt{3}}=\frac{l+m+n}{\sqrt{3}}$

Similarly,
$\cos b=\frac{-l+m+n}{\sqrt{3}}\cos c=\frac{l-m+n}{\sqrt{3}}\cos d=\frac{l+m-n}{\sqrt{3}}$

Squaring and adding
${\cos }^{2}a+{\cos }^{2}b+{\cos }^{2}c+{\cos }^{2}d=\frac{1}{3}\left[\right(l+m+n{)}^2+(l+m-n{)}^2+(l-m+n{)}^2+(-l+m+n{)}^2]=\frac{4}{3}(l^2+m^2+n^2)=\frac{4}{3}$