Let `veca =hati + hatj + sqrt2 hatk, vecb = b_(1) hati + b _(2) hatj + sqrt2 hatk and vecc = 5 hati + hatj + sqrt2 hatk` be three vectors such that the projection vector of `vecb` on `veca` is `veca.`
If `veca + vecb` perpendicular to `vecc,` then `|vecb|` is equal to

Answer

To solve the problem, we need to find the magnitude of the vector \(\vec{b}\) given the conditions involving the vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\).

### Step 1: Define the vectors
We have:
\[ \vec{a} = \hat{i} + \hat{j} + \sqrt{2} \hat{k} \]
\[ \vec{b} = b_1 \hat{i} + b_2 \hat{j} + \sqrt{2} \hat{k} \]
\[ \vec{c} = 5 \hat{i} + \hat{j} + \sqrt{2} \hat{k} \]

### Step 2: Find the magnitude of \(\vec{a}\)
The magnitude of \(\vec{a}\) is calculated as follows:
\[ |\vec{a}| = \sqrt{1^2 + 1^2 + (\sqrt{2})^2} = \sqrt{1 + 1 + 2} = \sqrt{4} = 2 \]

### Step 3: Use the projection formula
The projection of \(\vec{b}\) on \(\vec{a}\) is given by:
\[ \text{proj}_{\vec{a}} \vec{b} = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}|^2} \vec{a} \]
Given that this projection equals \(\vec{a}\), we have:
\[ \frac{\vec{a} \cdot \vec{b}}{|\vec{a}|^2} \vec{a} = \vec{a} \]
This implies:
\[ \vec{a} \cdot \vec{b} = |\vec{a}|^2 \]
Substituting the values:
\[ \vec{a} \cdot \vec{b} = (1)(b_1) + (1)(b_2) + (\sqrt{2})(\sqrt{2}) = b_1 + b_2 + 2 \]
Thus:
\[ b_1 + b_2 + 2 = 4 \quad \Rightarrow \quad b_1 + b_2 = 2 \quad \text{(Equation 1)} \]

### Step 4: Use the perpendicular condition
Since \(\vec{a} + \vec{b}\) is perpendicular to \(\vec{c}\), we have:
\[ (\vec{a} + \vec{b}) \cdot \vec{c} = 0 \]
Calculating \(\vec{a} + \vec{b}\):
\[ \vec{a} + \vec{b} = (1 + b_1) \hat{i} + (1 + b_2) \hat{j} + (2\sqrt{2}) \hat{k} \]
Now we compute the dot product:
\[ (1 + b_1) \cdot 5 + (1 + b_2) \cdot 1 + (2\sqrt{2}) \cdot \sqrt{2} = 0 \]
This simplifies to:
\[ 5 + 5b_1 + 1 + b_2 + 4 = 0 \]
Combining terms gives:
\[ 5b_1 + b_2 + 10 = 0 \quad \Rightarrow \quad 5b_1 + b_2 = -10 \quad \text{(Equation 2)} \]

### Step 5: Solve the system of equations
Now we have two equations:
1. \(b_1 + b_2 = 2\)
2. \(5b_1 + b_2 = -10\)

Subtract Equation 1 from Equation 2:
\[ (5b_1 + b_2) - (b_1 + b_2) = -10 - 2 \]
This simplifies to:
\[ 4b_1 = -12 \quad \Rightarrow \quad b_1 = -3 \]
Substituting \(b_1\) back into Equation 1:
\[ -3 + b_2 = 2 \quad \Rightarrow \quad b_2 = 5 \]

### Step 6: Write \(\vec{b}\) and find its magnitude
Now we can express \(\vec{b}\):
\[ \vec{b} = -3 \hat{i} + 5 \hat{j} + \sqrt{2} \hat{k} \]
To find the magnitude of \(\vec{b}\):
\[ |\vec{b}| = \sqrt{(-3)^2 + 5^2 + (\sqrt{2})^2} = \sqrt{9 + 25 + 2} = \sqrt{36} = 6 \]

### Final Answer
Thus, the magnitude of \(\vec{b}\) is:
\[ |\vec{b}| = 6 \]