If `vec(a)=3hat(i)+hat(j)-4hat(k), vec(b)=6hat(i)+5hat(j)-2hat(k)` and `|vec(c )|=3`, find the vector `vec(c )`, which is perpendicular to both `vec(a)` and `vec(b)`.

Answer

To find the vector \(\vec{c}\) that is perpendicular to both \(\vec{a}\) and \(\vec{b}\) and has a magnitude of 3, we can follow these steps:

### Step 1: Define the vectors
Given:
\[ \vec{a} = 3\hat{i} + \hat{j} - 4\hat{k} \]
\[ \vec{b} = 6\hat{i} + 5\hat{j} - 2\hat{k} \]

### Step 2: Calculate the cross product \(\vec{a} \times \vec{b}\)
The vector that is perpendicular to both \(\vec{a}\) and \(\vec{b}\) can be found using the cross product:
\[ \vec{d} = \vec{a} \times \vec{b} \]

To compute the cross product, we can use the determinant of a matrix:
\[ \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & -4 \\ 6 & 5 & -2 \end{vmatrix} \]

### Step 3: Calculate the determinant
Expanding the determinant:
\[ \vec{d} = \hat{i} \begin{vmatrix} 1 & -4 \\ 5 & -2 \end{vmatrix} - \hat{j} \begin{vmatrix} 3 & -4 \\ 6 & -2 \end{vmatrix} + \hat{k} \begin{vmatrix} 3 & 1 \\ 6 & 5 \end{vmatrix} \]

Calculating the 2x2 determinants:
1. For \(\hat{i}\):
\[ 1 \cdot (-2) - (-4) \cdot 5 = -2 + 20 = 18 \]
2. For \(\hat{j}\):
\[ 3 \cdot (-2) - (-4) \cdot 6 = -6 + 24 = 18 \quad \text{(remember to subtract this term)} \]
3. For \(\hat{k}\):
\[ 3 \cdot 5 - 1 \cdot 6 = 15 - 6 = 9 \]

Putting it all together:
\[ \vec{d} = 18\hat{i} - 18\hat{j} + 9\hat{k} \]

### Step 4: Find the magnitude of \(\vec{d}\)
Now, we compute the magnitude of \(\vec{d}\):
\[ |\vec{d}| = \sqrt{(18)^2 + (-18)^2 + (9)^2} = \sqrt{324 + 324 + 81} = \sqrt{729} = 27 \]

### Step 5: Find the unit vector in the direction of \(\vec{d}\)
The unit vector \(\hat{d}\) in the direction of \(\vec{d}\) is given by:
\[ \hat{d} = \frac{\vec{d}}{|\vec{d}|} = \frac{18\hat{i} - 18\hat{j} + 9\hat{k}}{27} = \frac{2}{3}\hat{i} - \frac{2}{3}\hat{j} + \frac{1}{3}\hat{k} \]

### Step 6: Scale the unit vector to the desired magnitude
To find \(\vec{c}\) with a magnitude of 3:
\[ \vec{c} = 3 \hat{d} = 3 \left(\frac{2}{3}\hat{i} - \frac{2}{3}\hat{j} + \frac{1}{3}\hat{k}\right) = 2\hat{i} - 2\hat{j} + \hat{k} \]

### Final Answer
Thus, the vector \(\vec{c}\) is:
\[ \vec{c} = 2\hat{i} - 2\hat{j} + \hat{k} \]