Case Study Questions on 3 D Geometry for Class 12

An aircraft navigation system is being designed to ensure safe flight paths over a mountainous region. A radar station is located at point \( P(1, 2, 3) \), and a signal beacon is at \( Q(4, 6, 7) \). The aircraft’s flight path is along a line passing through \( P \) with direction ratios \( (1, -2, 2) \). A restricted airspace plane is defined by \( 3x – y + 2z = 5 \). Engineers need to determine the direction cosines of the flight path, the equation of the path, the angle between the path and the line \( PQ \), the distance of the beacon from the plane, and conditions for coplanarity with another flight path to avoid collisions.

Key Formulas and Properties

• Direction Cosines: For a vector \( a\hat{i} + b\hat{j} + c\hat{k} \), direction cosines are \( \frac{a}{\sqrt{a^2 + b^2 + c^2}}, \frac{b}{\sqrt{a^2 + b^2 + c^2}}, \frac{c}{\sqrt{a^2 + b^2 + c^2}} \), and \( \cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1 \).
• Equation of a Line: Vector form: \( \vec{r} = \vec{a} + \lambda \vec{b} \); Cartesian form: \( \frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} \).
• Angle between Two Lines: For direction vectors \( (a_1, b_1, c_1) \) and \( (a_2, b_2, c_2) \), \( \cos\theta = \frac{a_1a_2 + b_1b_2 + c_1c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2} \cdot \sqrt{a_2^2 + b_2^2 + c_2^2}} \).
• Distance from a Point to a Plane: For point \( (x_1, y_1, z_1) \) and plane \( ax + by + cz + d = 0 \), distance = \( \left| \frac{ax_1 + by_1 + cz_1 + d}{\sqrt{a^2 + b^2 + c^2}} \right| \).
• Coplanarity Condition: Two lines with direction vectors \( \vec{b_1}, \vec{b_2} \) and a vector joining points on them \( \vec{a_2} – \vec{a_1} \) are coplanar if \( [\vec{a_2} – \vec{a_1}, \vec{b_1}, \vec{b_2}] = 0 \).

MCQ Questions

1. What are the direction cosines of the flight path with direction ratios \( (1, -2, 2) \)?
   \( \left( \frac{1}{3}, -\frac{2}{3}, \frac{2}{3} \right) \)

   \( \left( \frac{1}{\sqrt{9}}, -\frac{2}{\sqrt{9}}, \frac{2}{\sqrt{9}} \right) \)

   \( \left( \frac{1}{\sqrt{3}}, -\frac{2}{\sqrt{3}}, \frac{2}{\sqrt{3}} \right) \)

   \( \left( \frac{1}{3}, -\frac{1}{3}, \frac{2}{3} \right) \)

   Answer: (a) — The magnitude of the vector \( (1, -2, 2) \) is \( \sqrt{1^2 + (-2)^2 + 2^2} = 3 \). Direction cosines are \( \left(\frac{1}{3}, -\frac{2}{3}, \frac{2}{3}\right) \).

2. What is the Cartesian equation of the flight path passing through \( P(1, 2, 3) \) with direction ratios \( (1, -2, 2) \)?
   \( \frac{x-1}{1} = \frac{y-2}{-2} = \frac{z-3}{2} \)

   \( \frac{x-1}{2} = \frac{y-2}{-2} = \frac{z-3}{1} \)

   \( \frac{x-2}{1} = \frac{y-1}{-2} = \frac{z-3}{2} \)

   \( \frac{x-1}{1} = \frac{y-2}{2} = \frac{z-3}{-2} \)

   Answer: (a) — Line through \( P(1,2,3) \) with direction ratios \( (1,-2,2) \): \( \frac{x-1}{1} = \frac{y-2}{-2} = \frac{z-3}{2} \).

3. What is the distance from the beacon at \( Q(4, 6, 7) \) to the plane \( 3x – y + 2z = 5 \)?
   \( \frac{15}{\sqrt{14}} \)

   \( \frac{21}{\sqrt{14}} \)

   \( \frac{15}{\sqrt{13}} \)

   \( \frac{21}{\sqrt{13}} \)

   Answer: (a) — Distance \( = \left| \frac{3 \cdot 4 + (-1) \cdot 6 + 2 \cdot 7 – 5}{\sqrt{3^2 + (-1)^2 + 2^2}} \right| = \frac{15}{\sqrt{14}} \).

4. What is the angle between the flight path and the line joining \( P(1, 2, 3) \) to \( Q(4, 6, 7) \)?
   \( \cos^{-1}\left( \frac{3}{\sqrt{9} \cdot \sqrt{41}} \right) \)

   \( \cos^{-1}\left( \frac{9}{\sqrt{9} \cdot \sqrt{41}} \right) \)

   \( \cos^{-1}\left( \frac{3}{\sqrt{41} \cdot \sqrt{9}} \right) \)

   \( \cos^{-1}\left( \frac{9}{\sqrt{41} \cdot \sqrt{9}} \right) \)

   Answer: (b) — Using \( \cos\theta = \frac{1 \cdot 3 + (-2) \cdot 4 + 2 \cdot 4}{\sqrt{9} \cdot \sqrt{41}} = \frac{3 – 8 + 8}{3 \cdot \sqrt{41}} = \frac{3}{3 \cdot \sqrt{41}} = \frac{1}{\sqrt{41}} \), which matches option b.

5. What is the condition for the flight path to be coplanar with another line passing through \( Q(4, 6, 7) \) with direction ratios \( (2, 1, -1) \)?
   \( \begin{vmatrix} 3 & 4 & 4 \\ 1 & -2 & 2 \\ 2 & 1 & -1 \end{vmatrix} = 0 \)

   \( \begin{vmatrix} 1 & -2 & 2 \\ 3 & 4 & 4 \\ 2 & 1 & -1 \end{vmatrix} = 0 \)

   \( \begin{vmatrix} 4 & 6 & 7 \\ 1 & -2 & 2 \\ 2 & 1 & -1 \end{vmatrix} = 0 \)

   \( \begin{vmatrix} 1 & 2 & 3 \\ 4 & 6 & 7 \\ 2 & 1 & -1 \end{vmatrix} = 0 \)

   Answer: (a) — Coplanarity requires \( \begin{vmatrix} \vec{PQ} & \text{Flight path vector} & \text{New line vector} \end{vmatrix} = 0 \), where \( \vec{PQ} = (3,4,4) \).