Question

The equations of motion of a rocket are:
x = 2t,y = –4t, z = 4t, where the time t is given in seconds, and the coordinates of a ‘moving point in km. What is the path of the rocket? At what distances will the rocket be from the starting point O(0, 0, 0) and from the following line in 10 seconds? `vecr = 20hati - 10hatj + 40hatk + μ(10hati - 20hatj + 10hatk)`

Answer 1

Eliminating t between the equations, we obtain the equation of the path `x/2 = y/(-4) = z/4`, which is the equation of the line passing through the origin having direction ratios <2, –4, 4>. This line is the path of the rocket.

When t = 10 seconds, the rocket will be at the point (20, –40, 40).

Hence, the required distance from the origin at 10 seconds 

= `sqrt(20^2 + 40^2 + 40^2)` km

= 20 × 3 km

= 60 km

The distance of the point (20, –40, 40) from the given line

= `(|(veca_2 - veca_1) xx vecb|)/|vecb|`

= `(|-30hatj xx (10hati - 20hatj + 10hatk)|)/(|10hati - 20hatj + 10hatk|)` km

= `(|-300hati + 300hatk|)/(|10hati -  20hatj + 10hatk|)` km

= `(300sqrt(2))/(10sqrt(6))` km

= `10sqrt(3)` km