Find a vector of magnitude 9 units and perpendicular to the vectors
`veca=4hati-hatj+hatk and vecb= -2hati+hatj-2hatk`.

Answer

To find a vector of magnitude 9 units that is perpendicular to the vectors \(\vec{a} = 4\hat{i} - \hat{j} + \hat{k}\) and \(\vec{b} = -2\hat{i} + \hat{j} - 2\hat{k}\), we can follow these steps:

### Step 1: Find the cross product of \(\vec{a}\) and \(\vec{b}\)

The cross product \(\vec{a} \times \vec{b}\) gives us a vector that is perpendicular to both \(\vec{a}\) and \(\vec{b}\).

\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & -1 & 1 \\ -2 & 1 & -2 \end{vmatrix} \]

Calculating the determinant:

\[ \vec{a} \times \vec{b} = \hat{i} \begin{vmatrix} -1 & 1 \\ 1 & -2 \end{vmatrix} - \hat{j} \begin{vmatrix} 4 & 1 \\ -2 & -2 \end{vmatrix} + \hat{k} \begin{vmatrix} 4 & -1 \\ -2 & 1 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. \(\begin{vmatrix} -1 & 1 \\ 1 & -2 \end{vmatrix} = (-1)(-2) - (1)(1) = 2 - 1 = 1\)

2. \(\begin{vmatrix} 4 & 1 \\ -2 & -2 \end{vmatrix} = (4)(-2) - (1)(-2) = -8 + 2 = -6\)

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

Putting it all together:

\[ \vec{a} \times \vec{b} = \hat{i}(1) - \hat{j}(-6) + \hat{k}(2) = \hat{i} + 6\hat{j} + 2\hat{k} \]

### Step 2: Find the magnitude of \(\vec{a} \times \vec{b}\)

Now we calculate the magnitude of the vector \(\vec{c} = \hat{i} + 6\hat{j} + 2\hat{k}\):

\[ |\vec{c}| = \sqrt{(1)^2 + (6)^2 + (2)^2} = \sqrt{1 + 36 + 4} = \sqrt{41} \]

### Step 3: Normalize the vector \(\vec{c}\)

To find a unit vector in the direction of \(\vec{c}\):

\[ \hat{c} = \frac{\vec{c}}{|\vec{c}|} = \frac{\hat{i} + 6\hat{j} + 2\hat{k}}{\sqrt{41}} \]

### Step 4: Scale the unit vector to get a magnitude of 9

To find a vector of magnitude 9 that is perpendicular to both \(\vec{a}\) and \(\vec{b}\):

\[ \vec{d} = 9 \hat{c} = 9 \cdot \frac{\hat{i} + 6\hat{j} + 2\hat{k}}{\sqrt{41}} = \frac{9\hat{i} + 54\hat{j} + 18\hat{k}}{\sqrt{41}} \]

### Final Answer

The required vector of magnitude 9 units that is perpendicular to both \(\vec{a}\) and \(\vec{b}\) is:

\[ \vec{d} = \frac{9\hat{i} + 54\hat{j} + 18\hat{k}}{\sqrt{41}} \]