An airplane travels according to the vector $v=-20i+10j+3k$ and passes through the point $p_1=(1500,2000)$. There is a highway on the same co-ordinate system represented by the equation $2x+y=200$. At what height does the plane fly above the highway?
Solution : I believe here we are supposed to find the $y$- co-ordinate at which the line intersects the plane. Let's first write $v$ in parameter form : $$x=1500-20t$$$$y=2000+10t$$$$z=3t$$
Now plug these values in the equation of the plane $$2(1500-20t)+2000-10t=200$$$$t=160$$
This parameter gives us co-ordinates $$x=-1700$$$$y=3600$$$$z=480$$
These values satisfy the equation of the plane and therefore the height is 3600. Is this correct? Do we need to somehow consider the z co-ordinate as well?
The plane takes off from the point (1500, 2000) and into the direction (-20, 10, 3). The highway $2x+y=200$ is in the $xy$-plane and the $z$-direction represents the coordinate of the height.
Thus, the height for the plane to cross the highway from above is 480.