If `vec(A)=2hat(i)+3hat(j)+6hat(k)` and `vec(B)=3hat(i)-6hat(j)+2hat(k)` then vector perpendicular to both `vec(A)` and `vec(B)` has magnitude K times that of `6hat(i)+2hat(j)-3hat(k)`. Then K =

Options

A. 1
B. 3
C. 7
D. 9

Answer

To solve the problem, we need to find a vector that is perpendicular to both vectors \(\vec{A}\) and \(\vec{B}\), and then determine the value of \(K\) such that the magnitude of this perpendicular vector is \(K\) times the magnitude of another given vector.

### Step 1: Define the vectors
We are given:
\[ \vec{A} = 2\hat{i} + 3\hat{j} + 6\hat{k} \]
\[ \vec{B} = 3\hat{i} - 6\hat{j} + 2\hat{k} \]

### Step 2: Calculate the cross product \(\vec{A} \times \vec{B}\)
The vector perpendicular to both \(\vec{A}\) and \(\vec{B}\) can be found using the cross product:
\[ \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 6 \\ 3 & -6 & 2 \end{vmatrix} \]

Calculating the determinant:
\[ \vec{A} \times \vec{B} = \hat{i} \begin{vmatrix} 3 & 6 \\ -6 & 2 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 6 \\ 3 & 2 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & 3 \\ 3 & -6 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. For \(\hat{i}\):
\[ 3 \cdot 2 - (-6) \cdot 6 = 6 + 36 = 42 \]
2. For \(\hat{j}\):
\[ 2 \cdot 2 - 3 \cdot 6 = 4 - 18 = -14 \quad \text{(note the negative sign)} \]
3. For \(\hat{k}\):
\[ 2 \cdot (-6) - 3 \cdot 3 = -12 - 9 = -21 \]

Putting it all together:
\[ \vec{A} \times \vec{B} = 42\hat{i} + 14\hat{j} - 21\hat{k} \]

### Step 3: Calculate the magnitude of \(\vec{A} \times \vec{B}\)
The magnitude of the cross product is given by:
\[ |\vec{A} \times \vec{B}| = \sqrt{42^2 + 14^2 + (-21)^2} \]
Calculating each term:
- \(42^2 = 1764\)
- \(14^2 = 196\)
- \((-21)^2 = 441\)

Adding these:
\[ |\vec{A} \times \vec{B}| = \sqrt{1764 + 196 + 441} = \sqrt{2401} = 49 \]

### Step 4: Calculate the magnitude of the vector \(6\hat{i} + 2\hat{j} - 3\hat{k}\)
Now we calculate the magnitude of the vector:
\[ \vec{C} = 6\hat{i} + 2\hat{j} - 3\hat{k} \]
The magnitude is:
\[ |\vec{C}| = \sqrt{6^2 + 2^2 + (-3)^2} \]
Calculating each term:
- \(6^2 = 36\)
- \(2^2 = 4\)
- \((-3)^2 = 9\)

Adding these:
\[ |\vec{C}| = \sqrt{36 + 4 + 9} = \sqrt{49} = 7 \]

### Step 5: Find \(K\)
We know that the magnitude of \(\vec{A} \times \vec{B}\) is \(K\) times the magnitude of \(\vec{C}\):
\[ |\vec{A} \times \vec{B}| = K \cdot |\vec{C}| \]
Substituting the values we found:
\[ 49 = K \cdot 7 \]
Solving for \(K\):
\[ K = \frac{49}{7} = 7 \]

### Final Answer
Thus, the value of \(K\) is:
\[ \boxed{7} \]