If `vec(a)=2hat(i)-3hat(j)+4hat(k)` and `vec(b)=5hat(i)+hat(j)-hat(k)` represent sides of parallelogram, then find both diagonals and a unit vector perpendicular to both diagonals of parallelogram.

Answer

To solve the problem, we need to find the diagonals of the parallelogram formed by the vectors \(\vec{a}\) and \(\vec{b}\), and then find a unit vector that is perpendicular to both diagonals.

### Step 1: Identify the vectors
Given:
\[ \vec{a} = 2\hat{i} - 3\hat{j} + 4\hat{k} \]
\[ \vec{b} = 5\hat{i} + \hat{j} - \hat{k} \]

### Step 2: Find the first diagonal \(\vec{d_1}\)
The first diagonal of the parallelogram can be found by adding the two vectors:
\[ \vec{d_1} = \vec{a} + \vec{b} \]
Calculating this:
\[ \vec{d_1} = (2\hat{i} - 3\hat{j} + 4\hat{k}) + (5\hat{i} + \hat{j} - \hat{k}) \]
\[ = (2 + 5)\hat{i} + (-3 + 1)\hat{j} + (4 - 1)\hat{k} \]
\[ = 7\hat{i} - 2\hat{j} + 3\hat{k} \]

### Step 3: Find the second diagonal \(\vec{d_2}\)
The second diagonal can be found by subtracting vector \(\vec{b}\) from vector \(\vec{a}\):
\[ \vec{d_2} = \vec{a} - \vec{b} \]
Calculating this:
\[ \vec{d_2} = (2\hat{i} - 3\hat{j} + 4\hat{k}) - (5\hat{i} + \hat{j} - \hat{k}) \]
\[ = (2 - 5)\hat{i} + (-3 - 1)\hat{j} + (4 + 1)\hat{k} \]
\[ = -3\hat{i} - 4\hat{j} + 5\hat{k} \]

### Step 4: Find the cross product of the diagonals
To find a vector perpendicular to both diagonals, we calculate the cross product \(\vec{d_1} \times \vec{d_2}\):
\[ \vec{d_1} \times \vec{d_2} = (7\hat{i} - 2\hat{j} + 3\hat{k}) \times (-3\hat{i} - 4\hat{j} + 5\hat{k}) \]
Using the determinant method:
\[ \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 7 & -2 & 3 \\ -3 & -4 & 5 \end{vmatrix} \]

Calculating the determinant:
\[ = \hat{i} \begin{vmatrix} -2 & 3 \\ -4 & 5 \end{vmatrix} - \hat{j} \begin{vmatrix} 7 & 3 \\ -3 & 5 \end{vmatrix} + \hat{k} \begin{vmatrix} 7 & -2 \\ -3 & -4 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. \(\begin{vmatrix} -2 & 3 \\ -4 & 5 \end{vmatrix} = (-2)(5) - (3)(-4) = -10 + 12 = 2\)
2. \(\begin{vmatrix} 7 & 3 \\ -3 & 5 \end{vmatrix} = (7)(5) - (3)(-3) = 35 + 9 = 44\)
3. \(\begin{vmatrix} 7 & -2 \\ -3 & -4 \end{vmatrix} = (7)(-4) - (-2)(-3) = -28 - 6 = -34\)

Putting it all together:
\[ \vec{d_1} \times \vec{d_2} = 2\hat{i} - 44\hat{j} - 34\hat{k} \]

### Step 5: Find the magnitude of the cross product
Calculating the magnitude:
\[ |\vec{d_1} \times \vec{d_2}| = \sqrt{(2)^2 + (-44)^2 + (-34)^2} \]
\[ = \sqrt{4 + 1936 + 1156} = \sqrt{3096} \]

### Step 6: Find the unit vector perpendicular to both diagonals
The unit vector \(\hat{n}\) is given by:
\[ \hat{n} = \frac{\vec{d_1} \times \vec{d_2}}{|\vec{d_1} \times \vec{d_2}|} \]
Substituting the values:
\[ \hat{n} = \frac{2\hat{i} - 44\hat{j} - 34\hat{k}}{\sqrt{3096}} \]

### Final Answer
The diagonals of the parallelogram are:
1. \(\vec{d_1} = 7\hat{i} - 2\hat{j} + 3\hat{k}\)
2. \(\vec{d_2} = -3\hat{i} - 4\hat{j} + 5\hat{k}\)

The unit vector perpendicular to both diagonals is:
\[ \hat{n} = \frac{2\hat{i} - 44\hat{j} - 34\hat{k}}{\sqrt{3096}} \]