Let QQ be the cube with the set of vertices {(x_(1),x_(2),x_(3))in:}\left\{\left(x_{1}, x_{2}, x_{3}\right) \in\right.{:R^(3):x_(1),x_(2),x_(3)in{0,1}}\left.\mathbb{R}^{3}: x_{1}, x_{2}, x_{3} \in\{0,1\}\right\}
R.K. Malik's Newton Classes , Ranchi/Delhi
Let FF be the set of all twelve lines containing the diagonals of the six faces of the cube QQ. Let SS be the set of all four lines containing the main diagonals of the cube QQ; for instance, the line passing through the vertices (0,0,0)(0,0,0) and (1,1,1)(1,1,1) is in SS.
For lines ℓ_(1)\ell_{1} and ℓ_(2)\ell_{2}, let d(ℓ_(1),ℓ_(2))d\left(\ell_{1}, \ell_{2}\right) denote the shortest distance between them. Then the maximum value of d(ℓ_(1),ℓ_(2))d\left(\ell_{1}, \ell_{2}\right), as ℓ_(1)\ell_{1} varies over FF and ℓ_(2)\ell_{2} varies over SS, is
(A) (1)/(sqrt6)\frac{1}{\sqrt{6}}
(B) (1)/(sqrt8)\frac{1}{\sqrt{8}}
(C) (1)/(sqrt3)\frac{1}{\sqrt{3}}
(D) (1)/(sqrt12)\frac{1}{\sqrt{12}}
As every face has Two Diagonal Hence in Total there are 12 diagonals and FF is the set of all the diagonals
and SS is the set of body Diggonals like OAO A or EBE B etc.
R.K. Malik's Newton Classes , Ranchi/Delhi
By symmetry it is clear that shortest distance between any body diagoual and face diagonal is either 0 or some non-zero coustant
Hence we find the distance between oA and BD clearly d.r of OAO A are (1,1,1)(1,1,1) Hence the EqE q. of OAO A is (x-1)/(1)=(y-1)/(1)=(z-1)/(1)\frac{x-1}{1}=\frac{y-1}{1}=\frac{z-1}{1}
and d.r of BDB D are (1,0,-1)(1,0,-1)
{:[" Hence the "E_(q)" of "BD" is "(x-1)/(1)=(y-1)/(0)=(z)/(-1)],[S*D=(|[1-1,1-1,1-0],[1,1,1],[1,0,-1]|)/(sqrt((1-0)^(2)+(-1-1)^(2)+(0-1)^(2)))=(|[0,0,1],[1,1,1],[1,0,-1]|)/(sqrt(1+4+1))=(1)/(sqrt6)]:}\begin{aligned}
& \text { Hence the } E_{q} \text { of } B D \text { is } \frac{x-1}{1}=\frac{y-1}{0}=\frac{z}{-1} \\
& S \cdot D=\frac{\left|\begin{array}{ccc}
1-1 & 1-1 & 1-0 \\
1 & 1 & 1 \\
1 & 0 & -1
\end{array}\right|}{\sqrt{(1-0)^{2}+(-1-1)^{2}+(0-1)^{2}}}=\frac{\left|\begin{array}{ccc}
0 & 0 & 1 \\
1 & 1 & 1 \\
1 & 0 & -1
\end{array}\right|}{\sqrt{1+4+1}}=\frac{1}{\sqrt{6}}
\end{aligned}
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