Answer
The laser beam hits the $yz$-plane at the point $\left( {0,\frac{{11}}{2},38} \right)$.
Work Step by Step
We have ${\bf{r}}\left( t \right) = \left( { - \frac{{100}}{{{t^2}}},7 - t,40 - {t^2}} \right)$ for $0.5 \le t \le 1.5$. So, the velocity vector is
${\bf{r}}'\left( t \right) = \left( {\frac{{200}}{{{t^3}}}, - 1, - 2t} \right)$
Since the velocity vector is tangential to the path, when the laser is fired at $t=1$, its velocity:
${\bf{r}}'\left( 1 \right) = \left( {200, - 1, - 2} \right)$
is in the tangential direction of the laser moving toward the $yz$-plane.
At $t=1$, the spy plane is at the point ${\bf{r}}\left( t \right) = \left( { - 100,6,39} \right)$. So, the vector parametrization of the laser line is then
$\ell \left( u \right) = \left( { - 100,6,39} \right) + u\left( {200, - 1, - 2} \right)$ ${\ \ }$ for $0 \le u < \infty $
$\ell \left( u \right) = \left( { - 100 + 200u,6 - u,39 - 2u} \right)$
The laser beam hits the $yz$-plane when $x = - 100 + 200u = 0$. That is, when $u = \frac{1}{2}$.
Substituting it in $\ell \left( u \right)$ gives the point in the $yz$-plane: $\ell \left( {\frac{1}{2}} \right) = \left( {0,\frac{{11}}{2},38} \right)$. Thus, the laser beam hits the $yz$-plane at the point $\left( {0,\frac{{11}}{2},38} \right)$.
