If the magnitude of the vector product of the vector `hati+hatj+hatk` with a unit vector along the sum of vector `2hati+4hatj-5hatk` and `lambda hati+2hatj+3hatk` is equal to `sqrt2`, then find the value of `'lambda'`

Answer

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

### Step-by-Step Solution:

1. **Define the Vectors:**
Let \( \mathbf{A} = \hat{i} + \hat{j} + \hat{k} \) (Vector A).
Let \( \mathbf{B} = 2\hat{i} + 4\hat{j} - 5\hat{k} \) (Vector B).
Let \( \mathbf{C} = \lambda \hat{i} + 2\hat{j} + 3\hat{k} \) (Vector C).

2. **Calculate the Sum of Vectors B and C:**
\[ \mathbf{B} + \mathbf{C} = (2 + \lambda)\hat{i} + (4 + 2)\hat{j} + (-5 + 3)\hat{k} = (2 + \lambda)\hat{i} + 6\hat{j} - 2\hat{k} \]

3. **Find the Magnitude of the Sum:**
The magnitude of \( \mathbf{B} + \mathbf{C} \) is given by:
\[ |\mathbf{B} + \mathbf{C}| = \sqrt{(2 + \lambda)^2 + 6^2 + (-2)^2} \]
Simplifying this:
\[ |\mathbf{B} + \mathbf{C}| = \sqrt{(2 + \lambda)^2 + 36 + 4} = \sqrt{(2 + \lambda)^2 + 40} \]

4. **Calculate the Unit Vector:**
The unit vector along \( \mathbf{B} + \mathbf{C} \) is:
\[ \hat{u} = \frac{\mathbf{B} + \mathbf{C}}{|\mathbf{B} + \mathbf{C}|} = \frac{(2 + \lambda)\hat{i} + 6\hat{j} - 2\hat{k}}{\sqrt{(2 + \lambda)^2 + 40}} \]

5. **Find the Vector Product:**
Now we need to calculate the vector product \( \mathbf{A} \times \hat{u} \):
\[ \mathbf{A} \times \hat{u} = \left(\hat{i} + \hat{j} + \hat{k}\right) \times \left(\frac{(2 + \lambda)\hat{i} + 6\hat{j} - 2\hat{k}}{\sqrt{(2 + \lambda)^2 + 40}}\right) \]
Using the determinant method to calculate the cross product:
\[ \mathbf{A} \times \hat{u} = \frac{1}{\sqrt{(2 + \lambda)^2 + 40}} \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 2 + \lambda & 6 & -2 \end{vmatrix} \]

6. **Calculate the Determinant:**
Expanding the determinant:
\[ = \hat{i} \begin{vmatrix} 1 & 1 \\ 6 & -2 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 1 \\ 2 + \lambda & -2 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 1 \\ 2 + \lambda & 6 \end{vmatrix} \]
This results in:
\[ = \hat{i}(-2 - 6) - \hat{j}(-2 - (2 + \lambda)) + \hat{k}(6 - (2 + \lambda)) \]
Simplifying gives:
\[ = -8\hat{i} + (4 + \lambda)\hat{j} + (4 - \lambda)\hat{k} \]

7. **Magnitude of the Vector Product:**
The magnitude of the vector product is:
\[ |\mathbf{A} \times \hat{u}| = \frac{\sqrt{(-8)^2 + (4 + \lambda)^2 + (4 - \lambda)^2}}{\sqrt{(2 + \lambda)^2 + 40}} \]
Simplifying the numerator:
\[ = \sqrt{64 + (4 + \lambda)^2 + (4 - \lambda)^2} \]
Expanding gives:
\[ = \sqrt{64 + (16 + 8\lambda + \lambda^2) + (16 - 8\lambda + \lambda^2)} = \sqrt{64 + 32 + 2\lambda^2} = \sqrt{96 + 2\lambda^2} \]

8. **Setting the Magnitude Equal to \( \sqrt{2} \):**
We set the magnitude equal to \( \sqrt{2} \):
\[ \frac{\sqrt{96 + 2\lambda^2}}{\sqrt{(2 + \lambda)^2 + 40}} = \sqrt{2} \]
Squaring both sides:
\[ \frac{96 + 2\lambda^2}{(2 + \lambda)^2 + 40} = 2 \]
Cross-multiplying gives:
\[ 96 + 2\lambda^2 = 2((2 + \lambda)^2 + 40) \]
Expanding and simplifying leads to:
\[ 96 + 2\lambda^2 = 2(4 + 4\lambda + \lambda^2 + 40) = 2(44 + 4\lambda + \lambda^2) \]
\[ 96 + 2\lambda^2 = 88 + 8\lambda + 2\lambda^2 \]
Simplifying gives:
\[ 96 - 88 = 8\lambda \implies 8 = 8\lambda \implies \lambda = 1 \]

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