Adjacent sides of a parallelogram are given by vectors `2hat(i)+hat(j)+hat(k)` and `hat(i)+5hat(j)+hat(k)`. Find a unit vector in the direction of its diagonal. Also, find the area of parallelogram.

Answer

To solve the problem step-by-step, we will find a unit vector in the direction of the diagonal of the parallelogram formed by the given vectors and also calculate the area of the parallelogram.

### Given Vectors:
1. **Vector A (DA)**: \( \vec{A} = 2\hat{i} + \hat{j} + \hat{k} \)
2. **Vector B (DC)**: \( \vec{B} = \hat{i} + 5\hat{j} + \hat{k} \)

### Step 1: Find the Diagonal Vector
The diagonal vector \( \vec{D} \) from point D to point B can be found by adding the two adjacent vectors:

\[ \vec{D} = \vec{A} + \vec{B} \]

Substituting the values of \( \vec{A} \) and \( \vec{B} \):

\[ \vec{D} = (2\hat{i} + \hat{j} + \hat{k}) + (\hat{i} + 5\hat{j} + \hat{k}) \]

Now, combine the components:

\[ \vec{D} = (2 + 1)\hat{i} + (1 + 5)\hat{j} + (1 + 1)\hat{k} = 3\hat{i} + 6\hat{j} + 2\hat{k} \]

### Step 2: Find the Magnitude of the Diagonal Vector
The magnitude of the diagonal vector \( \vec{D} \) is given by:

\[ |\vec{D}| = \sqrt{(3)^2 + (6)^2 + (2)^2} \]

Calculating the squares:

\[ |\vec{D}| = \sqrt{9 + 36 + 4} = \sqrt{49} = 7 \]

### Step 3: Find the Unit Vector in the Direction of the Diagonal
The unit vector \( \hat{u} \) in the direction of the diagonal is given by:

\[ \hat{u} = \frac{\vec{D}}{|\vec{D}|} \]

Substituting the values:

\[ \hat{u} = \frac{3\hat{i} + 6\hat{j} + 2\hat{k}}{7} = \frac{3}{7}\hat{i} + \frac{6}{7}\hat{j} + \frac{2}{7}\hat{k} \]

### Step 4: Find the Area of the Parallelogram
The area \( A \) of the parallelogram formed by vectors \( \vec{A} \) and \( \vec{B} \) is given by the magnitude of the cross product of the two vectors:

\[ A = |\vec{A} \times \vec{B}| \]

### Step 5: Compute the Cross Product
To compute \( \vec{A} \times \vec{B} \):

\[ \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 1 \\ 1 & 5 & 1 \end{vmatrix} \]

Calculating the determinant:

\[ = \hat{i} \begin{vmatrix} 1 & 1 \\ 5 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 1 \\ 1 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & 1 \\ 1 & 5 \end{vmatrix} \]

Calculating the 2x2 determinants:

1. \( \begin{vmatrix} 1 & 1 \\ 5 & 1 \end{vmatrix} = (1)(1) - (1)(5) = 1 - 5 = -4 \)
2. \( \begin{vmatrix} 2 & 1 \\ 1 & 1 \end{vmatrix} = (2)(1) - (1)(1) = 2 - 1 = 1 \)
3. \( \begin{vmatrix} 2 & 1 \\ 1 & 5 \end{vmatrix} = (2)(5) - (1)(1) = 10 - 1 = 9 \)

Substituting back into the equation:

\[ \vec{A} \times \vec{B} = -4\hat{i} - 1\hat{j} + 9\hat{k} \]

### Step 6: Find the Magnitude of the Cross Product
Now, we find the magnitude:

\[ |\vec{A} \times \vec{B}| = \sqrt{(-4)^2 + (-1)^2 + (9)^2} = \sqrt{16 + 1 + 81} = \sqrt{98} = 7\sqrt{2} \]

### Final Answers
1. **Unit vector in the direction of the diagonal**:
\[ \hat{u} = \frac{3}{7}\hat{i} + \frac{6}{7}\hat{j} + \frac{2}{7}\hat{k} \]

2. **Area of the parallelogram**:
\[ A = 7\sqrt{2} \text{ square units} \]