If ` vec a , vec b` ,and `vec c` are there mutually perpendicular unit vectors and ` vec d` is a unit vector which makes equal angles with ` vec a , vec b` ,and `vec c`, then find the value of `| vec a+ vec b+ vec c+ vec d|^2`

Options

A. `4 + 2 sqrt(2)`
B. `4 + 2 sqrt(3)`
C. `2 + sqrt(5)`
D. `3 + sqrt(5)`

Answer

To solve the problem, we need to find the value of \( |\vec{a} + \vec{b} + \vec{c} + \vec{d}|^2 \) given that \( \vec{a}, \vec{b}, \) and \( \vec{c} \) are mutually perpendicular unit vectors, and \( \vec{d} \) is a unit vector that makes equal angles with \( \vec{a}, \vec{b}, \) and \( \vec{c} \).

### Step 1: Write the expression for the magnitude squared
We start with the expression:
\[ |\vec{a} + \vec{b} + \vec{c} + \vec{d}|^2 \]
Using the property of magnitudes, we can expand this as:
\[ |\vec{a} + \vec{b} + \vec{c} + \vec{d}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + |\vec{d}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} + \vec{a} \cdot \vec{d} + \vec{b} \cdot \vec{d} + \vec{c} \cdot \vec{d}) \]

### Step 2: Substitute known values
Since \( \vec{a}, \vec{b}, \) and \( \vec{c} \) are unit vectors:
\[ |\vec{a}|^2 = |\vec{b}|^2 = |\vec{c}|^2 = |\vec{d}|^2 = 1 \]
Thus, we have:
\[ |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + |\vec{d}|^2 = 1 + 1 + 1 + 1 = 4 \]

### Step 3: Evaluate the dot products
Since \( \vec{a}, \vec{b}, \) and \( \vec{c} \) are mutually perpendicular, we have:
\[ \vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{c} = \vec{c} \cdot \vec{a} = 0 \]
The remaining dot products involve \( \vec{d} \):
Let \( \vec{d} \) make an angle \( \theta \) with each of \( \vec{a}, \vec{b}, \vec{c} \). Then:
\[ \vec{d} \cdot \vec{a} = |\vec{d}||\vec{a}|\cos\theta = \cos\theta \]
\[ \vec{d} \cdot \vec{b} = \cos\theta \]
\[ \vec{d} \cdot \vec{c} = \cos\theta \]
Thus, we can write:
\[ \vec{a} \cdot \vec{d} + \vec{b} \cdot \vec{d} + \vec{c} \cdot \vec{d} = 3\cos\theta \]

### Step 4: Combine everything
Now substituting back into our expression:
\[ |\vec{a} + \vec{b} + \vec{c} + \vec{d}|^2 = 4 + 2(0 + 3\cos\theta) = 4 + 6\cos\theta \]

### Step 5: Find \( \cos\theta \)
Since \( \vec{d} \) is a unit vector making equal angles with \( \vec{a}, \vec{b}, \vec{c} \), we can express:
\[ \cos^2\theta + \cos^2\theta + \cos^2\theta = 1 \implies 3\cos^2\theta = 1 \implies \cos^2\theta = \frac{1}{3} \implies \cos\theta = \frac{1}{\sqrt{3}} \]

### Step 6: Substitute \( \cos\theta \) back
Now substituting \( \cos\theta \) into our expression:
\[ |\vec{a} + \vec{b} + \vec{c} + \vec{d}|^2 = 4 + 6\left(\frac{1}{\sqrt{3}}\right) = 4 + \frac{6}{\sqrt{3}} = 4 + 2\sqrt{3} \]

### Final Answer
Thus, the value of \( |\vec{a} + \vec{b} + \vec{c} + \vec{d}|^2 \) is:
\[ \boxed{4 + 2\sqrt{3}} \]