Question

Let a unit vector \( \hat{u} = x\hat{i} + y\hat{j} + z\hat{k} \) make angles \( \frac{\pi}{2}, \frac{\pi}{3} \), and \( \frac{2\pi}{3} \) with the vectors \( \frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{k} \), \( \frac{1}{\sqrt{2}} \hat{j} + \frac{1}{\sqrt{2}} \hat{k} \), and \( \frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{j} \) respectively. If \( \vec{v} = \frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{j} + \frac{1}{\sqrt{2}} \hat{k} \), then \( |\hat{u} - \vec{v}|^2 \) is equal to

• A. \( \frac{11}{2} \)
• B. \( \frac{5}{2} \)
• C. 9
• D. 7

Answer 1

The Correct Option is B

Given that \( \vec{u} = x\hat{i} + y\hat{j} + z\hat{k} \) is a unit vector, it satisfies: \(x^2 + y^2 + z^2 = 1\)
Step 1. Using the angle conditions:
  - The angle between \( \vec{u} \) and \( \frac{\hat{i}}{\sqrt{2}} + \frac{\hat{j}}{\sqrt{2}} \) is \( \frac{\pi}{2} \):  
   \(\vec{u} \cdot \left( \frac{\hat{i}}{\sqrt{2}} + \frac{\hat{j}}{\sqrt{2}} \right) = 0 \implies \frac{x}{\sqrt{2}} + \frac{y}{\sqrt{2}} = 0\)
   \(x + y = 0\) ---(1)
   - The angle between \( \vec{u} \) and \( \frac{\hat{i}}{\sqrt{2}} + \hat{j} + \frac{\hat{k}}{\sqrt{2}} \) is \( \frac{\pi}{3} \):  
    \(\vec{u} \cdot \left( \frac{\hat{i}}{\sqrt{2}} + \hat{j} + \frac{\hat{k}}{\sqrt{2}} \right) = \frac{1}{2} \implies \frac{x}{\sqrt{2}} + y + \frac{z}{\sqrt{2}} = \frac{1}{2}\)
   \(x + \sqrt{2}y + z = \frac{\sqrt{2}}{2}\)
    - The angle between \( \vec{u} \) and \( \frac{\hat{i}}{\sqrt{2}} + \frac{\hat{j}}{\sqrt{2}} + \hat{k} \) is \( \frac{\pi}{2} \):  
    \(\vec{u} \cdot \left( \frac{\hat{i}}{\sqrt{2}} + \frac{\hat{j}}{\sqrt{2}} + \hat{k} \right) = 0 \implies \frac{x}{\sqrt{2}} + \frac{y}{\sqrt{2}} + z = 0\)
    \(x + y + \sqrt{2}z = 0\)
 Step 2. Solving the system of equations:** From equations (1), (2), and (3):  
  - Substitute \( z = -x \) in (2):  
    \(x + \sqrt{2}y - x = \frac{\sqrt{2}}{2} \implies y = \frac{1}{\sqrt{2}}\)
   - Substitute \( y = \frac{1}{\sqrt{2}} \) and \( z = -x \) in (3):  
   \(x + \frac{1}{\sqrt{2}} + \sqrt{2}(-x) = 0 \implies x = -\frac{1}{2\sqrt{2}}, \, z = \frac{1}{2\sqrt{2}}\)

Step 3. Calculate \( |\vec{u} - \vec{v}|^2 \):
  \(|\vec{u} - \vec{v}|^2 = \left( x - \frac{1}{\sqrt{2}} \right)^2 + \left( y - \frac{1}{\sqrt{2}} \right)^2 + \left( z - \frac{1}{\sqrt{2}} \right)^2\)
  Substituting \( x = -\frac{1}{2\sqrt{2}}, \, y = \frac{1}{\sqrt{2}}, \, z = \frac{1}{2\sqrt{2}} \):  
  \(|\vec{u} - \vec{v}|^2 = \frac{5}{2}\)
The Correct answer is :\( \frac{5}{2} \).