Question

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.

Based on the above answer the following:

1. What is the path of the missile?

a. Straight line

b. Parabola

c. Circle

d. Ellipse

2. Which of the following points lie on the path of the missile?

a. (6, 8, 2)

b. (6, -8, -2)

c. (6, -8, 2)

d. (-6, -8, 2)

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

a. \(\sqrt{550}\) kms

b. \(\sqrt{650}\) kms

c. \(\sqrt{450}\) kms

d. \(\sqrt{750}\) kms

4. 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).

a. 12 km

b. 11 km

c. 20 km

d. 10 km

5. 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 the image of the position of the rocket in the sea is

Answer 1

1. (a) Straight line

2. (c) (6, -8, 2

3. (b) \(\sqrt{650}\) kms

4. (d) 10 km

5. (a) ( \(\cfrac{-9}7,\cfrac{-1}7\cfrac{-10}7\) )

Answer 2

1. (b) Straight line

Given equations are \(\frac x3 = \frac y{-4} =\frac z1\) is an equation of a straight line.

2. (c) (6, -8, 2)

Point (6, −8, 2) satisfy the equation of straight line.

3. (b) \(\sqrt{650}\) kms

In 5sec,

x = 15 km

y = −20 km

z = 5 km

Hence distance (0, 0, 0)(15, −20, 5) is \(\sqrt{650}\) kms.

4. (d) 10 km

5. (a) \(\left(\frac {-9}7, \frac {-1}7, \frac{-10}7\right)\)

Let image point (x₃, y₃, z₃) then by using

\(\frac{x_3 -x_1}{a} = \frac{y_3 - y_1}{b} = \frac{z_3 - z_1}{c}\)

\(= \frac{-2(ax_1 + by_1 + cx_1 + d)}{a^2 + b^2 + c^2}\)

Points are \(\left(\frac {-9}7, \frac {-1}7, \frac{-10}7\right)\).