$[$Size of face$].$ Let us calculate why our nose becomes bigger when focal length is small, as stated in the Lecture Note $4.$ Consider the following diagram of a simplified face setup, where we use a rectangle for our nose and a trapezoid of our face. The $3$D coordinates of $6$ points marked on our face are $A(-4,0,0),B(-2,0,-4),C(-1,0,-6),D(1,0,-6),E(2,0,-4),F(4,0,0).$ Their projection on the $2D$ image plane are $A^{\text{'}},B^{\text{'}},C^{\text{'}},D^{\text{'}},E^{\text{'}},$ and $F^{\text{'}}.$ Therefore, the width of our face on the $2D$ image is $A^{\text{'}}{F}^{\text{'}},$ the width of our nose is $C^{\text{'}}{D}^{\text{'}},$ and the width of one side face is $A^{\text{'}}{B}^{\text{'}}.$ Use this information to calculate:

a$)$ If we put a camera at $(0,0,-10)($so it is only $4$ away from the tip of our nose$),$ and take an image using a focal length equal to $5,$ what is the ratio between the width of our nose and our entire face $(C^{\text{'}}\frac{D^{\text{'}}}{A^{\text{'}}}{F}^{\text{'}})?$ What is the ratio between the one side face and our entire face $(\{$:$A^{\text{'}}\frac{B^{\text{'}}}{A^{\text{'}}}{F}^{\text{'}})?$

b$)$ Try to calculate two ratios again a$),$ assuming the camera is at $(0,0,-20($so it is $14$ away from the tip of the nose$),$ and focal length is $10.$

c$)$ Try to calculate the two ratios in a$)$ again using parallel projection and camera center is still at $(0,0,-30).$