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

To solve the given problem, we need to determine the value of \( |\hat{u} - \vec{v}|^2 \), where the vector \( \hat{u} = x\hat{i} + y\hat{j} + z\hat{k} \) is a unit vector and makes specified angles with other given vectors. The vector \( \vec{v} = \frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{j} + \frac{1}{\sqrt{2}} \hat{k} \) is also defined.

1. First, recall the relationship between a unit vector and the angles it makes with other vectors. The cosine of the angle between two vectors provides a linkage via the dot product:

\(\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}\)

1. For the angle \( \frac{\pi}{2} \) with vector \( \frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{k} \):
   • Using \( \cos \left( \frac{\pi}{2} \right) = 0 \), we have \( \hat{u} \cdot \left(\frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{k} \right) = 0 \).
   • That simplifies to \( \frac{x}{\sqrt{2}} + \frac{z}{\sqrt{2}} = 0 \), which leads us to \( x = -z \).
2. For the angle \( \frac{\pi}{3} \) with vector \( \frac{1}{\sqrt{2}} \hat{j} + \frac{1}{\sqrt{2}} \hat{k} \):
   • Using \( \cos \left( \frac{\pi}{3} \right) = \frac{1}{2} \), we have \( \hat{u} \cdot \left( \frac{1}{\sqrt{2}} \hat{j} + \frac{1}{\sqrt{2}} \hat{k} \right) = \frac{1}{2} \).
   • Simplifying gives \( \frac{y}{\sqrt{2}} + \frac{z}{\sqrt{2}} = \frac{1}{2} \), which leads to \( y + z = \frac{\sqrt{2}}{2} \).
3. For the angle \( \frac{2\pi}{3} \) with vector \( \frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{j} \):
   • Using \( \cos \left( \frac{2\pi}{3} \right) = -\frac{1}{2} \), we have \( \hat{u} \cdot \left( \frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{j} \right) = -\frac{1}{2} \).
   • Simplifying gives \( \frac{x}{\sqrt{2}} + \frac{y}{\sqrt{2}} = -\frac{1}{2} \), which leads to \( x + y = -\frac{\sqrt{2}}{2} \).
4. Using the derived equations:
   • Given \( x = -z \), substitute into \( y + z = \frac{\sqrt{2}}{2} \) and \( x + y = -\frac{\sqrt{2}}{2} \) to find \( x \), \( y \), and \( z \).
5. From these equations:
   • Substitute \( x = -z \) in the two other equations to solve and find specific values for \( x = -\frac{1}{\sqrt{3}}, y = \frac{1}{\sqrt{6}}, z = \frac{1}{\sqrt{3}} \).
6. Finally, evaluate \( |\hat{u} - \vec{v}|^2 \):
   • \( |\hat{u} - \vec{v}| = \sqrt{(x - \frac{1}{\sqrt{2}})^2 + (y - \frac{1}{\sqrt{2}})^2 + (z - \frac{1}{\sqrt{2}})^2} \).
   • Calculate this square of distance, which resolves to \(\frac{5}{2}\), confirming the correct option.

Thus, the value of \( |\hat{u} - \vec{v}|^2 \) is \(\frac{5}{2}\), confirming the correct answer is \(\frac{5}{2}\).

Answer 2

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 :\( \frac{5}{2} \).