Let `vec(A)=2hat(i)-3hat(j)+4hat(k)` and `vec(B)=4hat(i)+hat(j)+2hat(k)` then `|vec(A)xx vec(B)|` is equal to

Options

A. 440
B. `2sqrt(110)`
C. `sqrt(220)`
D. `4sqrt(65)`

Answer

To find the magnitude of the cross product of vectors \(\vec{A}\) and \(\vec{B}\), we will follow these steps:

### Step 1: Write down the vectors
Given:
\[ \vec{A} = 2\hat{i} - 3\hat{j} + 4\hat{k} \]
\[ \vec{B} = 4\hat{i} + \hat{j} + 2\hat{k} \]

### Step 2: Set up the determinant for the cross product
We will use the determinant method to calculate \(\vec{A} \times \vec{B}\). The determinant is set up as follows:
\[ \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -3 & 4 \\ 4 & 1 & 2 \end{vmatrix} \]

### Step 3: Calculate the determinant
To compute the determinant, we expand it:
\[ \vec{A} \times \vec{B} = \hat{i} \begin{vmatrix} -3 & 4 \\ 1 & 2 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 4 \\ 4 & 2 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & -3 \\ 4 & 1 \end{vmatrix} \]

Calculating each of these 2x2 determinants:
1. For \(\hat{i}\):
\[ \begin{vmatrix} -3 & 4 \\ 1 & 2 \end{vmatrix} = (-3)(2) - (4)(1) = -6 - 4 = -10 \]
2. For \(\hat{j}\):
\[ \begin{vmatrix} 2 & 4 \\ 4 & 2 \end{vmatrix} = (2)(2) - (4)(4) = 4 - 16 = -12 \]
3. For \(\hat{k}\):
\[ \begin{vmatrix} 2 & -3 \\ 4 & 1 \end{vmatrix} = (2)(1) - (-3)(4) = 2 + 12 = 14 \]

Putting it all together:
\[ \vec{A} \times \vec{B} = -10\hat{i} + 12\hat{j} + 14\hat{k} \]

### Step 4: Find the magnitude of the cross product
Now we calculate the magnitude:
\[ |\vec{A} \times \vec{B}| = \sqrt{(-10)^2 + (12)^2 + (14)^2} \]
Calculating each term:
\[ (-10)^2 = 100, \quad (12)^2 = 144, \quad (14)^2 = 196 \]
Adding these values:
\[ 100 + 144 + 196 = 440 \]
Thus,
\[ |\vec{A} \times \vec{B}| = \sqrt{440} \]

### Step 5: Simplify the square root
We can simplify \(\sqrt{440}\):
\[ \sqrt{440} = \sqrt{4 \times 110} = 2\sqrt{110} \]

### Final Answer
The magnitude of \(\vec{A} \times \vec{B}\) is:
\[ |\vec{A} \times \vec{B}| = 2\sqrt{110} \]

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