Unit Vector `hatu=xhati+yhatj+zhatk` make angles `pi/2, pi/3, (2pi)/3` with `(1/sqrt2hati+1/sqrt2hatk), (1/sqrt2hatj+1/sqrt2hatk)` and `(1/sqrt2hati+1/sqrt2hatj)` respectively and `vecv=1/sqrt2hati+1/sqrt2hatj+1/sqrt2hatk` Find `abs(vecu-vecv)`

Options

A. `sqrt(5/2)`
B. `sqrt(7/2)`
C. `sqrt(2/5)`
D. `sqrt(7/3)`

Answer

To solve the problem step by step, we need to find the unit vector \( \hat{u} = x \hat{i} + y \hat{j} + z \hat{k} \) that makes specific angles with given vectors. Then we will calculate the magnitude of the difference between \( \hat{u} \) and \( \vec{v} \).

### Step 1: Set up the vectors and angles
Given:
- \( \hat{u} = x \hat{i} + y \hat{j} + z \hat{k} \)
- \( \vec{v} = \frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{j} + \frac{1}{\sqrt{2}} \hat{k} \)

The angles are:
- \( \hat{u} \) makes an angle \( \frac{\pi}{2} \) with \( \vec{A} = \frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{k} \)
- \( \hat{u} \) makes an angle \( \frac{\pi}{3} \) with \( \vec{B} = \frac{1}{\sqrt{2}} \hat{j} + \frac{1}{\sqrt{2}} \hat{k} \)
- \( \hat{u} \) makes an angle \( \frac{2\pi}{3} \) with \( \vec{C} = \frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{j} \)

### Step 2: Use the dot product for angle calculations
1. For \( \hat{u} \) and \( \vec{A} \) (angle \( \frac{\pi}{2} \)):
\[ \hat{u} \cdot \vec{A} = 0 \]
\[ (x \hat{i} + y \hat{j} + z \hat{k}) \cdot \left(\frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{k}\right) = 0 \]
This gives:
\[ \frac{x}{\sqrt{2}} + \frac{z}{\sqrt{2}} = 0 \implies x + z = 0 \implies z = -x \]

2. For \( \hat{u} \) and \( \vec{B} \) (angle \( \frac{\pi}{3} \)):
\[ \hat{u} \cdot \vec{B} = |\hat{u}| |\vec{B}| \cos\left(\frac{\pi}{3}\right) \]
Since \( |\hat{u}| = 1 \) and \( |\vec{B}| = 1 \):
\[ \hat{u} \cdot \vec{B} = \frac{1}{2} \]
This gives:
\[ (x \hat{i} + y \hat{j} + z \hat{k}) \cdot \left(\frac{1}{\sqrt{2}} \hat{j} + \frac{1}{\sqrt{2}} \hat{k}\right) = \frac{1}{2} \]
Expanding this:
\[ \frac{y}{\sqrt{2}} + \frac{z}{\sqrt{2}} = \frac{1}{2} \implies y + z = \frac{\sqrt{2}}{2} \]

3. For \( \hat{u} \) and \( \vec{C} \) (angle \( \frac{2\pi}{3} \)):
\[ \hat{u} \cdot \vec{C} = |\hat{u}| |\vec{C}| \cos\left(\frac{2\pi}{3}\right) \]
This gives:
\[ (x \hat{i} + y \hat{j} + z \hat{k}) \cdot \left(\frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{j}\right) = -\frac{1}{2} \]
Expanding this:
\[ \frac{x}{\sqrt{2}} + \frac{y}{\sqrt{2}} = -\frac{1}{2} \implies x + y = -\frac{\sqrt{2}}{2} \]

### Step 3: Solve the equations
We have three equations:
1. \( z = -x \)
2. \( y + z = \frac{\sqrt{2}}{2} \)
3. \( x + y = -\frac{\sqrt{2}}{2} \)

Substituting \( z = -x \) into the second equation:
\[ y - x = \frac{\sqrt{2}}{2} \implies y = x + \frac{\sqrt{2}}{2} \]

Substituting \( y = x + \frac{\sqrt{2}}{2} \) into the third equation:
\[ x + \left(x + \frac{\sqrt{2}}{2}\right) = -\frac{\sqrt{2}}{2} \]
\[ 2x + \frac{\sqrt{2}}{2} = -\frac{\sqrt{2}}{2} \implies 2x = -\sqrt{2} \implies x = -\frac{\sqrt{2}}{2} \]

Then, substituting back to find \( y \) and \( z \):
\[ y = -\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = 0 \]
\[ z = -(-\frac{\sqrt{2}}{2}) = \frac{\sqrt{2}}{2} \]

Thus, we have:
\[ \hat{u} = -\frac{\sqrt{2}}{2} \hat{i} + 0 \hat{j} + \frac{\sqrt{2}}{2} \hat{k} \]

### Step 4: Calculate \( |\hat{u} - \vec{v}| \)
Now, we calculate:
\[ \hat{u} - \vec{v} = \left(-\frac{\sqrt{2}}{2} \hat{i} + 0 \hat{j} + \frac{\sqrt{2}}{2} \hat{k}\right) - \left(\frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{j} + \frac{1}{\sqrt{2}} \hat{k}\right) \]
\[ = \left(-\frac{\sqrt{2}}{2} - \frac{1}{\sqrt{2}}\right) \hat{i} + \left(0 - \frac{1}{\sqrt{2}}\right) \hat{j} + \left(\frac{\sqrt{2}}{2} - \frac{1}{\sqrt{2}}\right) \hat{k} \]

Calculating each component:
- For \( \hat{i} \):
\[ -\frac{\sqrt{2}}{2} - \frac{1}{\sqrt{2}} = -\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = -\sqrt{2} \]
- For \( \hat{j} \):
\[ 0 - \frac{1}{\sqrt{2}} = -\frac{1}{\sqrt{2}} \]
- For \( \hat{k} \):
\[ \frac{\sqrt{2}}{2} - \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = 0 \]

Thus:
\[ \hat{u} - \vec{v} = -\sqrt{2} \hat{i} - \frac{1}{\sqrt{2}} \hat{j} \]

### Step 5: Calculate the magnitude
Now, we find the magnitude:
\[ |\hat{u} - \vec{v}| = \sqrt{(-\sqrt{2})^2 + \left(-\frac{1}{\sqrt{2}}\right)^2 + 0^2} \]
\[ = \sqrt{2 + \frac{1}{2}} = \sqrt{\frac{5}{2}} = \frac{\sqrt{5}}{\sqrt{2}} = \frac{\sqrt{10}}{2} \]

### Final Answer
Thus, the final answer is:
\[ |\hat{u} - \vec{v}| = \frac{\sqrt{10}}{2} \]