Case-Study 2: Read the following passage and answer the questions given below.
The equation of motion of a missile are x = 3t, y = − 4t, z = t, where the time ‘t’ is given in

seconds,and the distance is measured in kilometres. 

i. At what distance will the rocket be from the starting point (0, 0, 0) in 5 s? 

ii. If the position of rocket at a certain instant of time is (5, - 8, 10), then what will be the height of the rocket from the ground? (The ground is considered as the xy-plane).

 iii. At a certain instant of time, if the missile is above the sea level, where the equation of the surface of sea is given by 2x + y + 3z =1 and the position of the missile at that instant of time is (1, 1, 2), then find the image of the position of the rocket in the sea.

i.  $$\begin{gathered} \mathrm{In} 5 \mathrm{s}, \mathrm{x}=15, \mathrm{y}=-20 \mathrm{and} \mathrm{z}=5 \\ \therefore \mathrm{required} \mathrm{distance}=\sqrt{{\left(15\right)}^2+{(-20)}^2+{\left(5\right)}^2} \\ =\sqrt{225+400+25}=\sqrt{650} \mathrm{km} \end{gathered}$$

ii. Since, the positive of rocket at a certain time is (5, -8, 10).
Height of the rocket from the ground (xy-plane) isz-coordinate of position of rocket i.e. 10 km.

iii. The equation of plane is

    $2\mathrm{x}+\mathrm{y}+3\mathrm{z}=1 \dots \left(\mathrm{i}\right)$$\therefore$ DR's of normal to plane (i) are (2, 1, 3)

$\therefore$ Equation of normal passing through the point (1, 1, 2) is

General point on the line is

$(2\mathrm{\lambda }+1, \mathrm{\lambda }+1, 3\mathrm{\lambda }+2)$

This point lies on the plane

$$\begin{gathered} 2(2\mathrm{\lambda }+1)+(\mathrm{\lambda }+1)+3(3\mathrm{\lambda }+2)=1 \\ \Rightarrow 4\mathrm{\lambda }+2+\mathrm{\lambda }+1+9\mathrm{\lambda }+6=1 \\ \Rightarrow 14\mathrm{\lambda }+9=1 \\ \Rightarrow 14\mathrm{\lambda }=-8 \\ \Rightarrow \mathrm{\lambda }=-\frac{4}{7} \\ \therefore \mathrm{Coordinate} \mathrm{of} \mathrm{the} \mathrm{foot} \mathrm{of} \mathrm{the} \mathrm{perpendicular} \mathrm{are} \\ \left(\frac{-8}{7}+1, \frac{-4}{7}+1, \frac{-12}{7}+2\right) \mathrm{i}.\mathrm{e}.\left(\frac{-1}{7}, \frac{3}{7}, \frac{2}{7}\right) \\ \mathrm{Let} (\mathrm{x}, \mathrm{y}, \mathrm{z}) \mathrm{be} \mathrm{the} \mathrm{image} \mathrm{of} (1, 1, 2) \mathrm{in} \mathrm{the} \mathrm{givin} \mathrm{plane} \\ \therefore \frac{1+\mathrm{x}}{2}=\frac{-1}{7}, \frac{1+}{2}=\frac{3}{7}\mathrm{and}\frac{2+\mathrm{z}}{2}=\frac{2}{7} \\ \Rightarrow \mathrm{x}=\frac{-9}{7}, \mathrm{y}=\frac{-1}{7}\mathrm{and} \mathrm{z}=\frac{-10}{7} \\ \therefore \mathrm{Required} \mathrm{image} \mathrm{is} \left(\frac{-9}{7}, \frac{-1}{7}, \frac{-10}{7}\right) \end{gathered}$$