The new position of $O,$ when triangle is rotated about side $AB$ by $90^\circ$ can be

Question

Consider the triangle $AOB$ in the $xy$-plane where $A\equiv(1,0,0);B\equiv(0,2,0);$ and $O(0,0,0)$.The new position of $O,$ when triangle is rotated about side $AB$ by $90^\circ$ can be
$(A)(\frac{4}{5},\frac{3}{5},\frac{2}{\sqrt5})\hspace{1cm}(B)(\frac{-3}{5},\frac{\sqrt2}{5},\frac{2}{\sqrt5})\hspace{1cm}(C)(\frac{4}{5},\frac{2}{5},\frac{2}{\sqrt5})\hspace{1cm}(D)(\frac{4}{5},\frac{3}{5},\frac{1}{\sqrt5})$

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I rotated the line $AB$ by hinging it at $B\equiv(0,2,0)$ and let the new position of $A$ be $A'(x,y,z)$
Now $AB$ is perpendicular to $A'B$.So sum of the product of their direction ratios is zero.
$x.1+(y-2).0+(z-0).0=0\Rightarrow x=0$
I think my approach is not right.Can someone please tell me the right approach to solve it.

Answer 1

Let $O'$ be the point we want. Also, let $C$ be a point on $AB$ such that $OC$ is perpendicular to $AB$.

Then, note that the coordinate of $O'$ is$$(C_x,C_y,\text{the length of the line segment $OC$})$$ where $C_x,C_y$ represents the $x$, $y$ coordinate of $C$ respectively.