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

Options

A. 6
B. 4
C. `sqrt(22)`
D. `sqrt(32)`

Answer

To solve the problem step by step, we will follow the instructions provided in the video transcript.

### Step 1: Write down the given vectors
We have three vectors:
- \( \vec{a} = \hat{i} + \hat{j} + \sqrt{2} \hat{k} \)
- \( \vec{b} = b_1 \hat{i} + b_2 \hat{j} + \sqrt{2} \hat{k} \)
- \( \vec{c} = 5 \hat{i} + \hat{j} + \sqrt{2} \hat{k} \)

### Step 2: Find the magnitude of vector \( \vec{a} \)
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 condition
The projection of \( \vec{b} \) on \( \vec{a} \) is given to be equal to \( |\vec{a}| \). The formula for the projection of \( \vec{b} \) on \( \vec{a} \) is:
\[ \text{Projection of } \vec{b} \text{ on } \vec{a} = \frac{\vec{b} \cdot \vec{a}}{|\vec{a}|} \]
Setting this equal to \( |\vec{a}| \) gives us:
\[ \frac{\vec{b} \cdot \vec{a}}{|\vec{a}|} = |\vec{a}| \]
This implies:
\[ \vec{b} \cdot \vec{a} = |\vec{a}|^2 = 2^2 = 4 \]

### Step 4: Calculate \( \vec{b} \cdot \vec{a} \)
Now we calculate \( \vec{b} \cdot \vec{a} \):
\[ \vec{b} \cdot \vec{a} = (b_1 \hat{i} + b_2 \hat{j} + \sqrt{2} \hat{k}) \cdot (\hat{i} + \hat{j} + \sqrt{2} \hat{k}) = b_1 \cdot 1 + b_2 \cdot 1 + \sqrt{2} \cdot \sqrt{2} \]
This simplifies to:
\[ b_1 + b_2 + 2 = 4 \]
Thus, we have:
\[ b_1 + b_2 = 2 \quad \text{(Equation 1)} \]

### Step 5: Use the perpendicularity condition
Next, we know that \( \vec{a} + \vec{b} \) is perpendicular to \( \vec{c} \). This means:
\[ (\vec{a} + \vec{b}) \cdot \vec{c} = 0 \]
Expanding this gives:
\[ \vec{a} \cdot \vec{c} + \vec{b} \cdot \vec{c} = 0 \]

### Step 6: Calculate \( \vec{a} \cdot \vec{c} \)
Calculating \( \vec{a} \cdot \vec{c} \):
\[ \vec{a} \cdot \vec{c} = (\hat{i} + \hat{j} + \sqrt{2} \hat{k}) \cdot (5 \hat{i} + \hat{j} + \sqrt{2} \hat{k}) = 5 + 1 + 2 = 8 \]

### Step 7: Substitute into the perpendicularity condition
Substituting into the perpendicularity condition gives:
\[ 8 + \vec{b} \cdot \vec{c} = 0 \implies \vec{b} \cdot \vec{c} = -8 \]

### Step 8: Calculate \( \vec{b} \cdot \vec{c} \)
Calculating \( \vec{b} \cdot \vec{c} \):
\[ \vec{b} \cdot \vec{c} = (b_1 \hat{i} + b_2 \hat{j} + \sqrt{2} \hat{k}) \cdot (5 \hat{i} + \hat{j} + \sqrt{2} \hat{k}) = 5b_1 + b_2 + 2 \]
Setting this equal to -8 gives:
\[ 5b_1 + b_2 + 2 = -8 \implies 5b_1 + b_2 = -10 \quad \text{(Equation 2)} \]

### Step 9: Solve the equations
Now we have two equations:
1. \( b_1 + b_2 = 2 \)
2. \( 5b_1 + b_2 = -10 \)

Subtracting Equation 1 from Equation 2:
\[ (5b_1 + b_2) - (b_1 + b_2) = -10 - 2 \]
This simplifies to:
\[ 4b_1 = -12 \implies b_1 = -3 \]

Substituting \( b_1 \) back into Equation 1:
\[ -3 + b_2 = 2 \implies b_2 = 5 \]

### Step 10: Find the magnitude of \( \vec{b} \)
Now we can write \( \vec{b} \):
\[ \vec{b} = -3 \hat{i} + 5 \hat{j} + \sqrt{2} \hat{k} \]
Calculating the magnitude:
\[ |\vec{b}| = \sqrt{(-3)^2 + 5^2 + (\sqrt{2})^2} = \sqrt{9 + 25 + 2} = \sqrt{36} = 6 \]

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