Case study Chapter 11(Three dimension geometry)

A line through the points A(3, 4, 1) and B(5, 1, 6) is drawn.(Case study three dimension geometry 2 )

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:(i)Answer (b)

Given points are A(3, 4, 1) and B(5, 1, 6)

∴ we have AB = $\sqrt{(5-3)^2+(1-4)^2+(6-1)^2}$

$=\sqrt{4+9+25} =\sqrt{38}$

∴ Its direction ratio are = $5-3, 1-4, 6-1$

= 2 , -3, 5

Direction cosine are = $\frac{2}{\sqrt{38}},\frac{-3}{\sqrt{38}},\frac{5}{\sqrt{38}}$

(ii) Answer (c)

Given points are A(3, 4, 1) and B(5, 1, 6)

∴ Direction given by = $5-3, 1-4, 6-1$

$= 2,-3,5$

(iii) Answer (a)

Equation of the line passing through the points are A(3, 4, 1) and B(5, 1, 6) in cartesian form is given by

$\frac{x-3}{5-3}=\frac{y-4}{1-4}=\frac{z-1}{6-1}$

i.e. $\frac{x-3}{2}=\frac{y-4}{-3}=\frac{z-1}{5}$

(iv) Answer (b)

Equation of the line passing through the points are A(3, 4, 1) and B(5, 1, 6) is

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

Since the line (i) croses the XY- plane

∴ The z co-ordinate of the point on the line (i) will be zero.

Putting z = 0 in (i), we get

⇒ $\frac{x-3}{2}=\frac{y-4}{-3}=\frac{0-1}{5}$

$\Rightarrow \frac{x-3}{2}=\frac{y-4}{-3}=\frac{-1}{5}$

⇒ $x = \frac{-2}{5}+3,y = \frac{3}{5}+4$

⇒ $x = \frac{13}{5}, y =\frac{23}{5}$

∴ Co-ordinates of the points is $(\frac{13}{5},\frac{23}{5},0)$

(v) Answer (c)

Equation of the line passing through the points are A(3, 4, 1) and B(5, 1, 6) is

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

Since the line (i) croses the XY- plane

∴ The y co-ordinate of the point on the line (i) will be zero.

Now, Putting z = 0 in (i), we get

⇒ $\frac{x-3}{2}=\frac{0-4}{-3}=\frac{z-1}{5}$

$\Rightarrow \frac{x-3}{2}=\frac{4}{3}=\frac{z-1}{5}$

⇒ $x = \frac{8}{3}+3,z = \frac{20}{3}+1$

⇒ $x = \frac{17}{3}, y =\frac{23}{3}$

∴ Co-ordinates of the points is $(\frac{17}{3}, 0, \frac{23}{3})$

Some other Case study

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 ?