Our goal is to create a perspective projection matrix. Recall its definition given in Lecture $9.$

$($a$)$ Give the projection matrix that results from a vertical field of view of $50,$ an aspect ratio of $2\mathrm{.}0,$ a near clipping plane of $1$ and far clipping plane of $10.$

Hint: consider that the field$-$of$-$view and the near clipping plane together create a right triangle. You can then use SOH$-$CAH$-$TOA to find the value of $t.$

Once you know $t,$ aspect ratio gives you $r.$

$($b$)$ To make sure you did this correctly, take the point $(0,0,-5\mathrm{.}5).$ If you multiply this point by your projection matrix, and then do perspective divide, you should get $(0,0,0),$ right? Indeed, $z=-5\mathrm{.}5$ is halfway between your near clipping plane and far clipping plane and therefore should be mapped to $z=0?$ No$,$ if you notice your entry in the projection matrix gives a nonlinear mapping. You'll get something close to $9/11-$