The equations of motion of a rocket are: $x = 2t, \quad y = -4t, \quad z = 4t,$ where the time $t$ is given in seconds, and the coordinates of a moving point are 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?

$\vec{r} = 20 \hat{i} - 10 \hat{j} + 40 \hat{k} + \mu (10 \hat{i} - 20 \hat{j} + 10 \hat{k})$

60 km, $10\sqrt{3}$ km

The correct answer is Option (1) → 60 km, $10\sqrt{3}$ km ##

Eliminating $t$ between the equations, we obtain the equation of the path $\frac{x}{2} = \frac{y}{-4} = \frac{z}{4},$ which are the equations of the line passing through the origin having direction ratios $\langle 2, -4, 4 \rangle$. 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 $= \sqrt{20^2 + (-40)^2 + 40^2} = 20 \times 3 = 60 \text{ km}$

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

$= \frac{|(\vec{a_2} - \vec{a_1}) \times \vec{b}|}{|\vec{b}|} = \frac{|(-30\hat{i}) \times (10\hat{i} - 20\hat{j} + 10\hat{k})|}{|10\hat{i} - 20\hat{j} + 10\hat{k}|} \text{ km}$

$= \frac{|-300\hat{i} + 300\hat{k}|}{|10\hat{i} - 20\hat{j} + 10\hat{k}|} = \frac{300\sqrt{2}}{10\sqrt{6}} = 10\sqrt{3} \text{ km}$