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 find the value of \( b \) such that the vectors \( \vec{P} = 2\hat{i} + b\hat{j} + 2\hat{k} \) and \( \vec{Q} = \hat{i} + \hat{j} + \hat{k} \) are perpendicular, we can follow these steps:

### Step 1: Understand the condition for perpendicular vectors
Two vectors are perpendicular if their dot product is equal to zero. Thus, we need to find \( \vec{P} \cdot \vec{Q} = 0 \).

**Hint:** Recall the formula for the dot product of two vectors.

### Step 2: Write the dot product of the vectors
The dot product \( \vec{P} \cdot \vec{Q} \) can be calculated as follows:
\[ \vec{P} \cdot \vec{Q} = (2\hat{i} + b\hat{j} + 2\hat{k}) \cdot (\hat{i} + \hat{j} + \hat{k}) \]

**Hint:** Use the distributive property of the dot product.

### Step 3: Calculate the dot product
Now, we compute the dot product:
\[ \vec{P} \cdot \vec{Q} = 2(\hat{i} \cdot \hat{i}) + b(\hat{j} \cdot \hat{j}) + 2(\hat{k} \cdot \hat{k}) + 2(\hat{i} \cdot \hat{j}) + b(\hat{j} \cdot \hat{i}) + 2(\hat{k} \cdot \hat{i}) + 2(\hat{i} \cdot \hat{k}) + b(\hat{j} \cdot \hat{k}) + 2(\hat{k} \cdot \hat{j}) \]
Since \( \hat{i} \cdot \hat{j} = 0 \), \( \hat{i} \cdot \hat{k} = 0 \), and \( \hat{j} \cdot \hat{k} = 0 \), we simplify to:
\[ \vec{P} \cdot \vec{Q} = 2(1) + b(1) + 2(1) = 2 + b + 2 = b + 4 \]

**Hint:** Remember that the dot product of the same unit vectors is 1.

### Step 4: Set the dot product equal to zero
Since the vectors are perpendicular, we set the dot product to zero:
\[ b + 4 = 0 \]

**Hint:** To isolate \( b \), you will need to perform basic algebra.

### Step 5: Solve for \( b \)
Now, solve for \( b \):
\[ b = -4 \]

**Hint:** Check if the value of \( b \) satisfies the condition of the problem.

### Conclusion
The value of \( b \) that makes the vectors \( \vec{P} \) and \( \vec{Q} \) perpendicular is:
\[ \boxed{-4} \]