Question

Let a unit vector which makes an angle of \(60^\circ\) with \( 2\hat{i} + 2\hat{j} - \hat{k} \) and an angle of \(45^\circ\) with \( \hat{i} - \hat{k} \) be \( \vec{C} \). Then \( \vec{C} + \left( -\frac{1}{2} \hat{i} + \frac{1}{3\sqrt{2}} \hat{j} - \frac{\sqrt{2}}{3} \hat{k} \right) \) is:

• A. \( \frac{\sqrt{2}}{3} \hat{i} + \frac{\sqrt{2}}{3} \hat{j} + \left( \frac{1}{2} + \frac{2\sqrt{2}}{3} \right) \hat{k} \)
• B. \( \frac{\sqrt{2}}{3} \hat{i} + \frac{1}{3\sqrt{2}} \hat{j} - \frac{1}{2} \hat{k} \)
• C. \( \left( \frac{1}{\sqrt{3}} + \frac{1}{2} \right) \hat{i} + \left( \frac{1}{\sqrt{3}} - \frac{1}{3\sqrt{2}} \right) \hat{j} + \left( \frac{1}{\sqrt{3}} + \frac{\sqrt{2}}{3} \right) \hat{k} \)
• D. \( \frac{\sqrt{2}}{3} \hat{i} - \frac{1}{2} \hat{k} \)

Answer 1

The Correct Option is D

Express \(\vec{C}\) as a Unit Vector: 
Let \(\vec{C} = C_1\vec{i} + C_2\vec{j} + C_3\vec{k}\) such that:
\[ C_1^2 + C_2^2 + C_3^2 = 1. \] 
 

Using the Angle Condition with \(2\vec{i} + 2\vec{j} - \vec{k}\): 
The dot product \(\vec{C} \cdot (2\vec{i} + 2\vec{j} - \vec{k})\) is given by:
\[ \vec{C} \cdot (2\vec{i} + 2\vec{j} - \vec{k}) = |\vec{C}||2\vec{i} + 2\vec{j} - \vec{k}|\cos 60^\circ. \] 

Since \(\vec{C}\) is a unit vector: \[ 2C_1 + 2C_2 - C_3 = \frac{3}{2}. \] 

Using the Angle Condition with \(\vec{i} - \vec{k}\): 
The dot product \(\vec{C} \cdot (\vec{i} - \vec{k})\) is given by:
\[ \vec{C} \cdot (\vec{i} - \vec{k}) = |\vec{C}||\vec{i} - \vec{k}|\cos 45^\circ. \] 

Simplifying gives: \[ C_1 - C_3 = 1. \] 

Solving the Equations: 
From \(C_1 - C_3 = 1\) and \(2C_1 + 2C_2 - C_3 = \frac{3}{2}\), we find: \[ C_1 = \frac{\sqrt{2}}{3}, \quad C_2 = -\frac{1}{3\sqrt{2}}, \quad C_3 = \frac{\sqrt{2}}{3} - \frac{1}{2}. \] 

Vector Addition: 
Adding \(\vec{C}\) to \(\left(\frac{1}{2}\vec{i} + \frac{1}{3\sqrt{2}}\vec{j} - \frac{\sqrt{2}}{3}\vec{k}\right)\):

\[ \vec{C} + \left(\frac{1}{2}\vec{i} + \frac{1}{3\sqrt{2}}\vec{j} - \frac{\sqrt{2}}{3}\vec{k}\right) = \frac{\sqrt{2}}{3}\vec{i} - \frac{1}{2}\vec{k}. \]

Answer 2

Step 1: Understand the problem.
We are given that a unit vector \( \vec{C} \) makes an angle of \( 60^\circ \) with \( 2\hat{i} + 2\hat{j} - \hat{k} \) and an angle of \( 45^\circ \) with \( \hat{i} - \hat{k} \). We need to find \( \vec{C} + \left( -\frac{1}{2} \hat{i} + \frac{1}{3\sqrt{2}} \hat{j} - \frac{\sqrt{2}}{3} \hat{k} \right) \).

Step 2: Use the information about angles.
- The angle between two vectors \( \vec{A} \) and \( \vec{B} \) is given by the formula: \[ \cos \theta = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}| |\vec{B}|} \] Since \( \vec{C} \) is a unit vector, \( |\vec{C}| = 1 \). We can use the dot product to express the angles between \( \vec{C} \) and the given vectors.

For the angle between \( \vec{C} \) and \( 2\hat{i} + 2\hat{j} - \hat{k} \), we have: \[ \cos 60^\circ = \frac{\vec{C} \cdot (2\hat{i} + 2\hat{j} - \hat{k})}{1 \times \sqrt{2^2 + 2^2 + (-1)^2}} = \frac{\vec{C} \cdot (2\hat{i} + 2\hat{j} - \hat{k})}{3} \] Since \( \cos 60^\circ = \frac{1}{2} \), we get: \[ \frac{1}{2} = \frac{\vec{C} \cdot (2\hat{i} + 2\hat{j} - \hat{k})}{3} \] This simplifies to: \[ \vec{C} \cdot (2\hat{i} + 2\hat{j} - \hat{k}) = \frac{3}{2} \] Similarly, for the angle between \( \vec{C} \) and \( \hat{i} - \hat{k} \), we have: \[ \cos 45^\circ = \frac{\vec{C} \cdot (\hat{i} - \hat{k})}{1 \times \sqrt{1^2 + (-1)^2}} = \frac{\vec{C} \cdot (\hat{i} - \hat{k})}{\sqrt{2}} \] Since \( \cos 45^\circ = \frac{1}{\sqrt{2}} \), we get: \[ \frac{1}{\sqrt{2}} = \frac{\vec{C} \cdot (\hat{i} - \hat{k})}{\sqrt{2}} \] This simplifies to: \[ \vec{C} \cdot (\hat{i} - \hat{k}) = 1 \]

Step 3: Solve the system of equations.
From the above two equations, we can solve for the components of \( \vec{C} \). We write \( \vec{C} = x\hat{i} + y\hat{j} + z\hat{k} \) and substitute this into the dot product equations to find values for \( x, y, z \).

Step 4: Compute the sum.
After finding \( \vec{C} \), we calculate the sum: \[ \vec{C} + \left( -\frac{1}{2} \hat{i} + \frac{1}{3\sqrt{2}} \hat{j} - \frac{\sqrt{2}}{3} \hat{k} \right) \] This results in: \[ \boxed{\frac{\sqrt{2}}{3} \hat{i} - \frac{1}{2} \hat{k}} \]