Let a unit vector `hat u = x hat i + y hat j + z hat k` make angles `(pi)2, (pi)/3` and `(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`

Options

A. `(11)/2`
B. `5/2`
C. 9
D. 7

Answer

To solve the problem, 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. We will derive equations based on the angles and then find the value of \(|\hat{u} - \vec{v}|^2\).

### Step 1: Set up the vectors and angles
We have:
- \(\vec{p_1} = \frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{k}\)
- \(\vec{p_2} = \frac{1}{\sqrt{2}} \hat{j} + \frac{1}{\sqrt{2}} \hat{k}\)
- \(\vec{p_3} = \frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{j}\)

The angles with \(\hat{u}\) are:
- \(\theta_1 = \frac{\pi}{2}\) with \(\vec{p_1}\)
- \(\theta_2 = \frac{\pi}{3}\) with \(\vec{p_2}\)
- \(\theta_3 = \frac{2\pi}{3}\) with \(\vec{p_3}\)

### Step 2: Use the dot product to derive equations
1. For \(\theta_1 = \frac{\pi}{2}\):
\[ \hat{u} \cdot \vec{p_1} = 0 \implies x \cdot \frac{1}{\sqrt{2}} + z \cdot \frac{1}{\sqrt{2}} = 0 \implies x + z = 0 \quad \text{(Equation 1)} \]

2. For \(\theta_2 = \frac{\pi}{3}\):
\[ \hat{u} \cdot \vec{p_2} = \frac{1}{2} \implies y \cdot \frac{1}{\sqrt{2}} + z \cdot \frac{1}{\sqrt{2}} = \frac{1}{2} \implies y + z = \frac{1}{\sqrt{2}} \quad \text{(Equation 2)} \]

3. For \(\theta_3 = \frac{2\pi}{3}\):
\[ \hat{u} \cdot \vec{p_3} = -\frac{1}{2} \implies x \cdot \frac{1}{\sqrt{2}} + y \cdot \frac{1}{\sqrt{2}} = -\frac{1}{2} \implies x + y = -\frac{1}{\sqrt{2}} \quad \text{(Equation 3)} \]

### Step 3: Solve the system of equations
From Equation 1:
\[ z = -x \]

Substituting \(z\) into Equation 2:
\[ y - x = \frac{1}{\sqrt{2}} \implies y = x + \frac{1}{\sqrt{2}} \quad \text{(Equation 4)} \]

Substituting \(y\) from Equation 4 into Equation 3:
\[ x + (x + \frac{1}{\sqrt{2}}) = -\frac{1}{\sqrt{2}} \implies 2x + \frac{1}{\sqrt{2}} = -\frac{1}{\sqrt{2}} \implies 2x = -\frac{2}{\sqrt{2}} \implies x = -\frac{1}{\sqrt{2}} \]

Using \(x\) to find \(y\) and \(z\):
\[ y = -\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = 0 \]
\[ z = -(-\frac{1}{\sqrt{2}}) = \frac{1}{\sqrt{2}} \]

Thus, we have:
\[ \hat{u} = -\frac{1}{\sqrt{2}} \hat{i} + 0 \hat{j} + \frac{1}{\sqrt{2}} \hat{k} \]

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

Now, calculate the magnitude squared:
\[ |\hat{u} - \vec{v}|^2 = \left(-\sqrt{2}\right)^2 + \left(-\frac{1}{\sqrt{2}}\right)^2 + 0^2 = 2 + \frac{1}{2} = \frac{5}{2} \]

### Final Answer
\[ |\hat{u} - \vec{v}|^2 = \frac{5}{2} \]