What is the value of b such that the scalar product of the vector `hat(i)+hat(j)+hat(k)` with the unit vector parallel to the sum of the vectors `2hat(i)+4hat(j)-5 hat(k) and b hat(i)+2hat(j)+3hat(k) ` is unity ?

Options

A. -2
B. -1
C. 0
D. 1

Explanation

To solve the problem, we need to find the value of \( b \) such that the scalar product of the vector \( \hat{i} + \hat{j} + \hat{k} \) with the unit vector parallel to the sum of the vectors \( 2\hat{i} + 4\hat{j} - 5\hat{k} \) and \( b\hat{i} + 2\hat{j} + 3\hat{k} \) is equal to 1.

### Step-by-step Solution:

1. **Define the vectors:**
Let:
\[ \mathbf{A} = \hat{i} + \hat{j} + \hat{k} \]
\[ \mathbf{B} = 2\hat{i} + 4\hat{j} - 5\hat{k} \]
\[ \mathbf{C} = b\hat{i} + 2\hat{j} + 3\hat{k} \]

2. **Find the sum of vectors \( \mathbf{B} \) and \( \mathbf{C} \):**
\[ \mathbf{B} + \mathbf{C} = (2 + b)\hat{i} + (4 + 2)\hat{j} + (-5 + 3)\hat{k} \]
Simplifying this gives:
\[ \mathbf{B} + \mathbf{C} = (2 + b)\hat{i} + 6\hat{j} - 2\hat{k} \]

3. **Find the magnitude of the resultant vector:**
The magnitude of \( \mathbf{B} + \mathbf{C} \) is:
\[ \|\mathbf{B} + \mathbf{C}\| = \sqrt{(2 + b)^2 + 6^2 + (-2)^2} \]
Calculating this gives:
\[ \|\mathbf{B} + \mathbf{C}\| = \sqrt{(2 + b)^2 + 36 + 4} = \sqrt{(2 + b)^2 + 40} \]

4. **Find the unit vector in the direction of \( \mathbf{B} + \mathbf{C} \):**
The unit vector \( \mathbf{N} \) parallel to \( \mathbf{B} + \mathbf{C} \) is given by:
\[ \mathbf{N} = \frac{\mathbf{B} + \mathbf{C}}{\|\mathbf{B} + \mathbf{C}\|} = \frac{(2 + b)\hat{i} + 6\hat{j} - 2\hat{k}}{\sqrt{(2 + b)^2 + 40}} \]

5. **Set up the scalar product equation:**
We need the scalar product \( \mathbf{A} \cdot \mathbf{N} = 1 \):
\[ (\hat{i} + \hat{j} + \hat{k}) \cdot \left( \frac{(2 + b)\hat{i} + 6\hat{j} - 2\hat{k}}{\sqrt{(2 + b)^2 + 40}} \right) = 1 \]
This simplifies to:
\[ \frac{(2 + b) + 6 - 2}{\sqrt{(2 + b)^2 + 40}} = 1 \]
Which can be rewritten as:
\[ \frac{(b + 6)}{\sqrt{(2 + b)^2 + 40}} = 1 \]

6. **Cross-multiply to eliminate the fraction:**
\[ b + 6 = \sqrt{(2 + b)^2 + 40} \]

7. **Square both sides:**
\[ (b + 6)^2 = (2 + b)^2 + 40 \]
Expanding both sides gives:
\[ b^2 + 12b + 36 = b^2 + 4b + 4 + 40 \]
Simplifying this results in:
\[ b^2 + 12b + 36 = b^2 + 4b + 44 \]

8. **Rearranging the equation:**
\[ 12b - 4b + 36 - 44 = 0 \]
This simplifies to:
\[ 8b - 8 = 0 \]
Thus:
\[ 8b = 8 \implies b = 1 \]

### Final Answer:
The value of \( b \) is \( 1 \).