If the extremities of a diagonal of a square are $(1,-2,3)$ and $(2,-3,5)$, then the length of its side is
If the extremities of a diagonal of a square are $(1,-2,3)$ and $(2,-3,5)$, then the length of its side is
$\sqrt{6}$
$\sqrt{3}$
$\sqrt{5}$
$\sqrt{7}$
Solution
Let a be the length of the square.
Let $B \equiv(1,-2,3)$ and $D \equiv(2,-3,5)$
$\begin{aligned}
& \therefore B D=\sqrt{(1-2)^2+(-2+3)^2+(3-5)^2} \\
& B D=\sqrt{1+1+4}=\sqrt{6} \\
& \text { In } \triangle A B D: A B^2+A D^2=B D^2 \\
& \Rightarrow a^2+a^2=6 \\
& \Rightarrow 2 a^2=6 \Rightarrow a^2=3 \\
& \Rightarrow a=\sqrt{3}
\end{aligned}$
