If `vec( a) = hat(i) + hat(j) + p hat(k)` and `vec( b) = vec( i) + hat(j) + hat(k)` then `| vec( a) + hat(b) | = | vec( a) |+ | vec( b)|` holds for

Options

A. all real p
B. no real p
C. p =-1
D. p =1

Answer

To solve the problem, we need to analyze the vectors given and check under what conditions the equation \( |\vec{a} + \vec{b}| = |\vec{a}| + |\vec{b}| \) holds true.

### Step 1: Define the vectors
Given:
\[ \vec{a} = \hat{i} + \hat{j} + p\hat{k} \]
\[ \vec{b} = \hat{i} + \hat{j} + \hat{k} \]

### Step 2: Calculate the magnitudes of the vectors
The magnitude of vector \(\vec{a}\) is:
\[ |\vec{a}| = \sqrt{(1)^2 + (1)^2 + (p)^2} = \sqrt{2 + p^2} \]

The magnitude of vector \(\vec{b}\) is:
\[ |\vec{b}| = \sqrt{(1)^2 + (1)^2 + (1)^2} = \sqrt{3} \]

### Step 3: Calculate the sum of the vectors
Now, we compute \(\vec{a} + \vec{b}\):
\[ \vec{a} + \vec{b} = (\hat{i} + \hat{j} + p\hat{k}) + (\hat{i} + \hat{j} + \hat{k}) = 2\hat{i} + 2\hat{j} + (p + 1)\hat{k} \]

### Step 4: Calculate the magnitude of the sum
Now, we find the magnitude of \(\vec{a} + \vec{b}\):
\[ |\vec{a} + \vec{b}| = \sqrt{(2)^2 + (2)^2 + (p + 1)^2} = \sqrt{4 + 4 + (p + 1)^2} = \sqrt{8 + (p + 1)^2} \]

### Step 5: Set up the equation
We need to check when:
\[ |\vec{a} + \vec{b}| = |\vec{a}| + |\vec{b}| \]
Substituting the magnitudes we calculated:
\[ \sqrt{8 + (p + 1)^2} = \sqrt{2 + p^2} + \sqrt{3} \]

### Step 6: Square both sides to eliminate the square root
Squaring both sides gives:
\[ 8 + (p + 1)^2 = (2 + p^2) + 3 + 2\sqrt{(2 + p^2)(3)} \]
This simplifies to:
\[ 8 + p^2 + 2p + 1 = 5 + p^2 + 2\sqrt{(2 + p^2)(3)} \]
\[ 9 + 2p = 5 + 2\sqrt{(2 + p^2)(3)} \]
\[ 4 + 2p = 2\sqrt{(2 + p^2)(3)} \]
Dividing by 2:
\[ 2 + p = \sqrt{(2 + p^2)(3)} \]

### Step 7: Square both sides again
Squaring both sides again gives:
\[ (2 + p)^2 = (2 + p^2)(3) \]
Expanding both sides:
\[ 4 + 4p + p^2 = 6 + 3p^2 \]
Rearranging gives:
\[ 0 = 2p^2 - 4p + 2 \]
Dividing by 2:
\[ 0 = p^2 - 2p + 1 \]
Factoring:
\[ (p - 1)^2 = 0 \]

### Step 8: Solve for \(p\)
Thus, we find:
\[ p = 1 \]

### Conclusion
The condition for which \( |\vec{a} + \vec{b}| = |\vec{a}| + |\vec{b}| \) holds is when \( p = 1 \).