Question

If a line makes $\alpha ,\beta ,\gamma ,\delta$ angles with 4 diagonals of a cube , then ${\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma +{\cos }^2\delta =?$

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

Answer 1

The correct option is B $\frac{2}{3}$

Take $O$ as a corner

$OA,OB,OC$ are $3$ edges thr0ugh the axes

Let $OA=OB=OC=a$

coordinates 0f $O=(0,0,0)$

$A(a,0,0),B(0,a,0),C(0,0,a)$

$P(a,a,0),L(0,a,a),M(a,0,a),N(a,a,0)$

The f0ur diag0nals $OP,AL,BM,CN$

Direction cosine Of $OP:a-0,a-0,a-0=a,a,a=1,1,1$

Direction cosine 0f $AL:0-a,a-0,a-0=-a,a,a=-1,1,1$

Direction cosine 0f $BM:a-0,0-a,a-0=a,-a,a=1,-1,1$

Direction cosine 0f $CN:a-0,a-0,0-a=a,a,-a=1,1,-1$

$\therefore$D.C's of $OP$ are $\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}$

D.C's of $AL$ are $-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}$

D.C's of $BM$ are $\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}$

D.C's of $CN$ are $\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}$

Let $l,m,n$ be dc's of line and line ma\dfrac{1}{\sqrt{3}}es angle $\alpha$

With $OP:\cos \alpha l\left(\frac{1}{\sqrt{3}}\right)+m\left(\frac{1}{\sqrt{3}}\right)+n\left(\frac{1}{\sqrt{3}}\right)=\frac{l+m+n}{\sqrt{3}}$

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

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

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

squaring and adding all the four

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

$=\frac{1}{3}[{(l+m+n)}^2+{(-l+m+n)}^2+{(l-m+n)}^2+{(l+m-n)}^2]$

$=\frac{1}{3}[4l^2+4m^2+4n^2]$

$=\frac{4}{3}$ since $l^2+m^2+n^2=1$

$\therefore {\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma +{\cos }^2\delta =\frac{4}{3}$