Chapter 13 - Vector Functions - 13.1 Exercises - Page 871: 47

Yes

$t^2-4t+3=0$ ...(1) $ \implies t^2--3t-t+3=0$ thus, $ t= 3$ $t^2-7t+12=0$ ...(2) $ \implies t^2-6t-t+12=0$ and $t= 3$ We need to set the $z$ component of the two equations equal to each other first. We have $t^2=5t-6$ ...(3) This implies that $t^2-5t+6=0$ and $t^2-6t+t+6=0$ Thus, we have $t=3$ Yes, all three equations give the same result.