Question

Three friends A, B, and C move out from the same location O at the same time in three different directions to reach their destinations. They move out on straight paths and decide that A and B after reaching their destinations will meet up with C at his pre-decided destination, following straight paths from A to C and B to C in such a way that \( \overrightarrow{OA} = \hat{i}, \overrightarrow{OB} = \hat{j} \), and \( \overrightarrow{OC} = 5 \hat{i} - 2 \hat{j} \), respectively.

Based upon the above information, answer the following questions: 
(i) Complete the given figure to explain their entire movement plan along the respective vectors.}
(ii) Find vectors \( \overrightarrow{AC} \) and \( \overrightarrow{BC} \).}
(iii) (a) If \( \overrightarrow{a} = 2 \hat{i} - \hat{j} + 4 \hat{k} \), distance of O to A is 1 km, and from O to B is 2 km, then find the angle between \( \overrightarrow{OA} \) and \( \overrightarrow{OB} \). Also, find \( | \overrightarrow{a} \times \overrightarrow{b} | \).}
(iii) (b) If \( \overrightarrow{a} = 2 \hat{i} - \hat{j} + 4 \hat{k} \), find a unit vector perpendicular to \( (\overrightarrow{a} + \overrightarrow{b}) \) and \( (\overrightarrow{a} - \overrightarrow{b}) \). 

Hint:

For cross products, remember the determinant formula and make sure to break down each component step-by-step. A perpendicular unit vector can be easily found by normalizing the cross product result.

Answer 1

(i) Complete the figure:

We are given:

• \( \vec{OA} = \vec{a} \)
• \( \vec{OB} = \vec{b} \)
• \( \vec{OC} = 5\vec{a} - 2\vec{b} \)

From this, we can compute:

\[ \vec{AC} = \vec{OC} - \vec{OA} = (5\vec{a} - 2\vec{b}) - \vec{a} = 4\vec{a} - 2\vec{b} \]
\[ \vec{BC} = \vec{OC} - \vec{OB} = (5\vec{a} - 2\vec{b}) - \vec{b} = 5\vec{a} - 3\vec{b} \]

(ii) Find Vectors \( \vec{AC} \) and \( \vec{BC} \)

\[ \vec{AC} = 4\vec{a} - 2\vec{b}, \quad \vec{BC} = 5\vec{a} - 3\vec{b} \]

(iii-a) Given \( \vec{a} \cdot \vec{b} = 1 \), \( |\vec{a}| = 1 \, \text{km} \), \( |\vec{b}| = 2 \, \text{km} \):

Step 1: Find angle between \( \vec{a} \) and \( \vec{b} \)

\[ \vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta \Rightarrow 1 = (1)(2)\cos\theta \Rightarrow \cos\theta = \frac{1}{2} \Rightarrow \theta = \cos^{-1}\left(\frac{1}{2}\right) = 60^\circ \]

Step 2: Find \( |\vec{a} \times \vec{b}| \)

\[ |\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin\theta = (1)(2)\sin(60^\circ) = 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3} \]

(iii-b) Given \( \vec{a} = 2\hat{i} - \hat{j} + 4\hat{k} \), \( \vec{b} = \hat{j} - \hat{k} \):

Step 1: Find \( \vec{a} + \vec{b} \)

\[ \vec{a} + \vec{b} = (2\hat{i} - \hat{j} + 4\hat{k}) + (\hat{j} - \hat{k}) = 2\hat{i} + 0\hat{j} + 3\hat{k} \]

Step 2: Find \( \vec{a} - \vec{b} \)

\[ \vec{a} - \vec{b} = (2\hat{i} - \hat{j} + 4\hat{k}) - (\hat{j} - \hat{k}) = 2\hat{i} - 2\hat{j} + 5\hat{k} \]

Step 3: Cross Product

\[ \vec{u} = (\vec{a} + \vec{b}) \times (\vec{a} - \vec{b}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 0 & 3 \\ 2 & -2 & 5 \\ \end{vmatrix} \] \[ = \hat{i}(0 \cdot 5 - 3 \cdot (-2)) - \hat{j}(2 \cdot 5 - 3 \cdot 2) + \hat{k}(2 \cdot (-2) - 0 \cdot 2) = \hat{i}(6) - \hat{j}(10 - 6) + \hat{k}(-4) = 6\hat{i} - 4\hat{j} - 4\hat{k} \]

Step 4: Unit Vector

\[ |\vec{u}| = \sqrt{6^2 + (-4)^2 + (-4)^2} = \sqrt{36 + 16 + 16} = \sqrt{68} \] \[ \text{Unit Vector} = \frac{1}{\sqrt{68}}(6\hat{i} - 4\hat{j} - 4\hat{k}) \]

Final Answers:

1. \( \vec{AC} = 4\vec{a} - 2\vec{b} \)
2. \( \vec{BC} = 5\vec{a} - 3\vec{b} \)
3. Angle between \( \vec{a} \) and \( \vec{b} \): \( 60^\circ \)
4. \( |\vec{a} \times \vec{b}| = \sqrt{3} \)
5. Unit vector = \( \frac{1}{\sqrt{68}}(6\hat{i} - 4\hat{j} - 4\hat{k}) \)