Mathematics - 3D-Coordinate Geometry Question with Solution | TestHub

length of projection of $AB$ in plane is $A^{\prime }B$ for ${\mathrm{A}}^{\prime }$ $\frac{x-2}{1}=\frac{y-3}{1}=\frac{z-1}{1}=-\left(\frac{2+3+1-3}{3}\right)$ $x-2=-1y-3=-1z-2=-1$ $\mathrm{x}=1\mathrm{y}=2\mathrm{z}=0$ ${\mathrm{A}}^{\prime }\equiv (1,2,0)$ ${\mathrm{A}}^{\prime }\mathrm{B}=\sqrt{(1-1{)}^2+(2-1{)}^2+(1-0{)}^2}$ $=\sqrt{2}$ \end{document}