Two vector `vec(P) = 2hat(i) + bhat(j) + 2hat(k)` and `vec(Q) = hat(i) + hat(j) + hat(k)` are perpendicular. The value of b will be :

Options

A. b = 0
B. b = 1
C. b = 2
D. b = -4

Answer

To solve the problem, we need to find the value of \( b \) such that the vectors \( \vec{P} \) and \( \vec{Q} \) are perpendicular.

### Step-by-Step Solution:

1. **Identify the Vectors**:
\[ \vec{P} = 2\hat{i} + b\hat{j} + 2\hat{k} \]
\[ \vec{Q} = \hat{i} + \hat{j} + \hat{k} \]

2. **Use the Condition for Perpendicular Vectors**:
Two vectors are perpendicular if their dot product is zero:
\[ \vec{P} \cdot \vec{Q} = 0 \]

3. **Calculate the Dot Product**:
The dot product \( \vec{P} \cdot \vec{Q} \) is calculated as follows:
\[ \vec{P} \cdot \vec{Q} = (2\hat{i} + b\hat{j} + 2\hat{k}) \cdot (\hat{i} + \hat{j} + \hat{k}) \]
This expands to:
\[ = 2(\hat{i} \cdot \hat{i}) + b(\hat{j} \cdot \hat{j}) + 2(\hat{k} \cdot \hat{k}) \]
Since \( \hat{i} \cdot \hat{i} = 1 \), \( \hat{j} \cdot \hat{j} = 1 \), and \( \hat{k} \cdot \hat{k} = 1 \), we have:
\[ = 2(1) + b(1) + 2(1) = 2 + b + 2 \]
Simplifying this gives:
\[ = 4 + b \]

4. **Set the Dot Product to Zero**:
Since the vectors are perpendicular:
\[ 4 + b = 0 \]

5. **Solve for \( b \)**:
Rearranging the equation gives:
\[ b = -4 \]

### Final Answer:
The value of \( b \) is \( -4 \).

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