For a rectangle base is `(hat(i) + hat(j) + hat(k))` and adjacent side is `(4 hat(j) + 3hat(k))`. If the area of rectangle is `5^(n)` then value of n is

Answer

To solve the problem, we need to find the area of a rectangle given its base and adjacent side vectors. Let's break down the steps:

### Step 1: Define the vectors
Let:
- **Base vector** \( \mathbf{a} = \hat{i} + \hat{j} + \hat{k} \)
- **Adjacent side vector** \( \mathbf{b} = 4\hat{j} + 3\hat{k} \)

### Step 2: Calculate the cross product \( \mathbf{a} \times \mathbf{b} \)
The area of the rectangle can be found using the magnitude of the cross product of the two vectors. We will use the determinant method to calculate \( \mathbf{a} \times \mathbf{b} \).

\[ \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 0 & 4 & 3 \end{vmatrix} \]

### Step 3: Expand the determinant
Using the determinant expansion, we have:

\[ \mathbf{a} \times \mathbf{b} = \hat{i} \begin{vmatrix} 1 & 1 \\ 4 & 3 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 1 \\ 0 & 3 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 1 \\ 0 & 4 \end{vmatrix} \]

Calculating the 2x2 determinants:

1. For \( \hat{i} \):
\[ \begin{vmatrix} 1 & 1 \\ 4 & 3 \end{vmatrix} = (1)(3) - (1)(4) = 3 - 4 = -1 \]

2. For \( \hat{j} \):
\[ \begin{vmatrix} 1 & 1 \\ 0 & 3 \end{vmatrix} = (1)(3) - (1)(0) = 3 - 0 = 3 \]

3. For \( \hat{k} \):
\[ \begin{vmatrix} 1 & 1 \\ 0 & 4 \end{vmatrix} = (1)(4) - (1)(0) = 4 - 0 = 4 \]

Putting it all together:

\[ \mathbf{a} \times \mathbf{b} = -\hat{i} - 3\hat{j} + 4\hat{k} \]

### Step 4: Calculate the magnitude of the cross product
Now, we find the magnitude of \( \mathbf{a} \times \mathbf{b} \):

\[ |\mathbf{a} \times \mathbf{b}| = \sqrt{(-1)^2 + (-3)^2 + (4)^2} = \sqrt{1 + 9 + 16} = \sqrt{26} \]

### Step 5: Set the area equal to \( 5^n \)
We know that the area of the rectangle is given as:

\[ \text{Area} = |\mathbf{a} \times \mathbf{b}| = \sqrt{26} \]

We are given that this area equals \( 5^n \):

\[ \sqrt{26} = 5^n \]

### Step 6: Solve for \( n \)
To solve for \( n \), we can take logarithms:

\[ \log(\sqrt{26}) = n \log(5) \]

This simplifies to:

\[ \frac{1}{2} \log(26) = n \log(5) \]

Thus, we can isolate \( n \):

\[ n = \frac{\frac{1}{2} \log(26)}{\log(5)} = \frac{\log(26)}{2 \log(5)} \]

### Conclusion
The value of \( n \) is:

\[ n = \frac{\log(26)}{2 \log(5)} \]