Question Number 6054 by Rasheed Soomro last updated on 11/Jun/16
$$\mathcal{D}{etermine}\:{distance}\:{between} \\ $$$${opposite}\:{corners}\:\:{of}\:{a}\:{cubic} \\ $$$${room}\:{of}\:{dimention}\:{x}\:{units}. \\ $$$$ \\ $$
Commented by Yozzii last updated on 11/Jun/16
$${length}\:{of}\:{diagonal}\:{on}\:{one}\:{face}={x}\sqrt{\mathrm{2}}\:{units} \\ $$$${By}\:{Phythagorean}\:{theorem},\:{required}\:{distance}=\sqrt{\left({x}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} +{x}^{\mathrm{2}} }={x}\sqrt{\mathrm{3}} \\ $$$$ \\ $$$${In}\:\mathrm{3}{D}\:{coordinates}\:{let}\:{one}\:{vertex}\:{be}\:{the}\:{origin} \\ $$$${and}\:{an}\:{opposite}\:{vertex}\:{be}\:\begin{pmatrix}{{x}}\\{{x}}\\{{x}}\end{pmatrix}. \\ $$$$\Rightarrow\mid\begin{pmatrix}{{x}}\\{{x}}\\{{x}}\end{pmatrix}\:\mid=\sqrt{{x}^{\mathrm{2}} +{x}^{\mathrm{2}} +{x}^{\mathrm{2}} }={x}\sqrt{\mathrm{3}}. \\ $$
Commented by Rasheed Soomro last updated on 11/Jun/16
$$\mathfrak{Th}^{\mathfrak{a}} \mathfrak{nkS}! \\ $$
Answered by Yozzii last updated on 11/Jun/16
$${x}\sqrt{\mathrm{3}} \\ $$