Case study three dimension geometry 1 chapter 11 class 12

Case study Chapter 11 (Three dimension geometry)

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.(Case study three dimension geometry 1)

Case study three dimension geometry 1 — The equation of motion of a missile are x = 3t, y = -4t, z = t where the time

(i) What is the path of the missile ?

(a) Straight line (b) Parabola

(ii) Which of the following points lie on the path of the missile ?

(a) (6, 8, 2) (b) (6, -8, -2)

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

(a) √550 kms (b) √650 kms

(iv) If the position of rocket at 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

(v) For what value of k are the lines
$\frac{x-1}{2}=\frac{y-1}{3}=\frac{z-1}{k}$ and $\frac{x-1}{-2}=\frac{y-3}{-1}=\frac{z-5}{7}$ perpendicular ?

(a) 1 (b) 2

Solution:(i) Answer (a)

Given equation of motion of a missile be

x = 3t, y = -4t, z = t

⇒ $\frac{x}{3} = \frac{y}{-4}=\frac{z}{1}$ ← which is a straight lne.

Hence, path of the missile is a straight line.

(ii) Answer (c)

We have equation of the path of the missile as

$\frac{x}{3} = \frac{y}{-4}=\frac{z}{1}$

and the point (6, -8, 2) satisfy the equation

∴ point is (6, -8, 2) lie on the path of missile.

(iii) Answer (b)

After 5 seconds positionof the rocket be

x = 3t = 3×5 = 15

y = -4t = -4×5 = -20

z = t = 5

∴ point is (15, -20, 5).

Its distance from origin (0, 0, 0) is

= $\sqrt{(15-0)^2+(-20-0)^2+(5-0)^2}$

$=\sqrt{225+400+25}=\sqrt{650}$ kms

(iv) Answer (d)

Given position of the rocket at a time is (5, -8, 10)

∴ Height of the rocket from the ground

=Distance between the points (5, -8, 10) and (5, -8, 0).

(Since, ground is considered as the XY-plane)

$=\sqrt{(5-5)^2+(-8+8)^2+(10-0)^2}=10$ km

(v) Answer (a)

$\frac{x-1}{2}=\frac{y-1}{3}=\frac{z-1}{k}$ and $\frac{x-1}{-2}=\frac{y-3}{-1}=\frac{z-5}{7}$

dr’s of first line (2, 3, k)

And dr’s of second line (-2, -1, 7)

Given lines are perpendicular if

⇒ 2×-2 + 3×-1 + k×7 = 0

⇒ -4 – 3 + 7k = 0

⇒ 7k = 7

⇒ k = 1

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Question: A line through the points A(3, 4, 1) and B(5, 1, 6) is drawn.

Based on the above information answer the following questions:

(i) Direction cosines of the line passing through the points A and B are

(a) $\left(\frac{2}{\sqrt{38}}, \frac{3}{\sqrt{38}}, \frac{5}{\sqrt{38}}\right)$

(b) $\left(\frac{2}{\sqrt{38}}, \frac{-3}{\sqrt{38}}, \frac{5}{\sqrt{38}}\right)$

(d) $\left(\frac{-2}{\sqrt{38}}, \frac{-3}{\sqrt{38}}, \frac{-5}{\sqrt{38}}\right)$

(ii) Direction ratios of the line passing through the points A and B are

(a) -2, -3, 5 (b) 2, 3, 5

(iii) Equation of the line passing through the points A and B in certain form is

(a) $\frac{x-3}{2}=\frac{y-4}{-3}=\frac{z-1}{5}$

(b) $\frac{x-3}{-2}=\frac{y-4}{-3}=\frac{z-1}{5}$

(d) $\frac{x-5}{-2}=\frac{y-1}{-3}=\frac{z-6}{-5}$

(iv) The coordinates of the point where the line through the points A and B crosses the XY-plane is

(a) $\left(\frac{-3}{2}, \frac{4}{3}, \frac{-1}{5}\right)$

(b) $\left(\frac{13}{5}, \frac{23}{5}, 0 \right)$

(d) $\left(\frac{17}{3}, \frac{23}{3}, 0\right)$

(v) The coordinates of the point where the line through the points A and B crosses the XZ-plane is

(a) $\left(\frac{13}{5}, 0, \frac{23}{5}\right)$

(b) $\left(\frac{-3}{2}, 0, \frac{-1}{5}\right)$

(d) $\left(\frac{17}{5}, \frac{23}{5}, 0\right)$

Solution: For solution click here

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