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 \(\left(\frac{1}{2}\vec{i} + \frac{1}{3\sqrt{2}}\vec{j} - \frac{\sqrt{2}}{3}\vec{k}\right)\):

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