Question

lf a line makes angles ${60}^o,{45}^o,{45}^o$ and $\theta$ with the four diagonals of a cube, then ${\sin }^2\theta =$

• A. $\frac{1}{12}$
• B. $\frac{11}{12}$
• C. $\sqrt{\frac{11}{12}}$
• D. $\frac{31}{12}$

Answer 1

Answer: (B)

$\frac{11}{12}$

Let the length of the cube be $a$ units.
Let $A{A}^{\prime },B{B}^{\prime },C{C}^{\prime },OP$ be the diagonals of the cube.
Coordinates of vertices are $O(0,0,0),A(a,0,0),A^{\prime }(0,a,a),B(0,b,0),B^{\prime }(b,0,b),C(0,0,c),C^{\prime }(c,c,0),P(a,a,a)$
D.Rs of $A{A}^{\prime }$ are $(-a,a,a)$.
D.Rs of $B{B}^{\prime }$ are $(a,-a,a)$
D.Rs of $C{C}^{\prime }$ are $(a,a,-a)$
D.Rs of $OP$ are $(a,a,a)$
D.c's of $A{A}^{\prime }$ are $\frac{-a}{\sqrt{a^2+a^2+a^2}},\frac{a}{\sqrt{a^2+a^2+a^2}},\frac{a}{\sqrt{a^2+a^2+a^2}}=-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}$
D.c's of $B{B}^{\prime }$ are $\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}$
D.c's of $C{C}^{\prime }$ are $\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}$
D.c's of $OP$ are $\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}$
Let the d.c's of given line be $l,m,n$ such that $l^2+m^2+n^2=1$
Given , the line makes angles ${60}^o,{45}^o,{45}^o$ and $\theta$ with the four diagonals of a cube.
$\cos {60}^0=\frac{-l+m+n}{\sqrt{3}}$
$\Rightarrow \frac{1}{2}=\frac{-l+m+n}{\sqrt{3}}$ .....(1)
$\cos {45}^0=\frac{l-m+n}{\sqrt{3}}$
$\Rightarrow \frac{1}{\sqrt{2}}=\frac{l-m+n}{\sqrt{3}}$ .....(2)
$\cos {45}^0=\frac{l+m-n}{\sqrt{3}}$
$\Rightarrow \frac{1}{\sqrt{2}}=\frac{l+m-n}{\sqrt{3}}$ .....(3)
$\cos \theta =\frac{l+m+n}{\sqrt{3}}$ ....(4)
Squaring and adding (1),(2),(3),(4), we get
$\Rightarrow \frac{5}{4}+{\cos }^2\theta =\frac{4}{3}$ ($\because l^2+m^2+n^2=1$)
$\Rightarrow {\cos }^2\theta =\frac{1}{12}$
$\Rightarrow {\sin }^2\theta =\frac{11}{12}$