If` vec(a) = 2hat(i) - 3 hat(j) + 4 hat (k), b = vec(i)+2vec (j) -hat(k),vec(c ) = 3 hat(i) - hat(j) -2 hat(k)` , then the volume ( in cubic units) of the parallelopiped with coterminous edges `vec(a) + vec(b) , vec(b) + vec(c ) , vec(c ) + vec( a) ` is

Options

A. 35
B. 0
C. 105
D. 70

Answer

To find the volume of the parallelepiped formed by the vectors \(\vec{a} + \vec{b}\), \(\vec{b} + \vec{c}\), and \(\vec{c} + \vec{a}\), we will follow these steps:

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

### Step 2: Calculate \(\vec{a} + \vec{b}\)
\[ \vec{a} + \vec{b} = (2\hat{i} - 3\hat{j} + 4\hat{k}) + (\hat{i} + 2\hat{j} - \hat{k}) \]
Combine the components:
\[ = (2 + 1)\hat{i} + (-3 + 2)\hat{j} + (4 - 1)\hat{k} = 3\hat{i} - \hat{j} + 3\hat{k} \]

### Step 3: Calculate \(\vec{b} + \vec{c}\)
\[ \vec{b} + \vec{c} = (\hat{i} + 2\hat{j} - \hat{k}) + (3\hat{i} - \hat{j} - 2\hat{k}) \]
Combine the components:
\[ = (1 + 3)\hat{i} + (2 - 1)\hat{j} + (-1 - 2)\hat{k} = 4\hat{i} + \hat{j} - 3\hat{k} \]

### Step 4: Calculate \(\vec{c} + \vec{a}\)
\[ \vec{c} + \vec{a} = (3\hat{i} - \hat{j} - 2\hat{k}) + (2\hat{i} - 3\hat{j} + 4\hat{k}) \]
Combine the components:
\[ = (3 + 2)\hat{i} + (-1 - 3)\hat{j} + (-2 + 4)\hat{k} = 5\hat{i} - 4\hat{j} + 2\hat{k} \]

### Step 5: Form the matrix for the volume
The volume \(V\) of the parallelepiped is given by the absolute value of the determinant of the matrix formed by the vectors:
\[ \begin{vmatrix} 3 & -1 & 3 \\ 4 & 1 & -3 \\ 5 & -4 & 2 \end{vmatrix} \]

### Step 6: Calculate the determinant
Using the determinant formula for a \(3 \times 3\) matrix:
\[ V = \begin{vmatrix} 3 & -1 & 3 \\ 4 & 1 & -3 \\ 5 & -4 & 2 \end{vmatrix} = 3 \begin{vmatrix} 1 & -3 \\ -4 & 2 \end{vmatrix} - (-1) \begin{vmatrix} 4 & -3 \\ 5 & 2 \end{vmatrix} + 3 \begin{vmatrix} 4 & 1 \\ 5 & -4 \end{vmatrix} \]

Calculating the \(2 \times 2\) determinants:
1. \(\begin{vmatrix} 1 & -3 \\ -4 & 2 \end{vmatrix} = (1)(2) - (-3)(-4) = 2 - 12 = -10\)

2. \(\begin{vmatrix} 4 & -3 \\ 5 & 2 \end{vmatrix} = (4)(2) - (-3)(5) = 8 + 15 = 23\)

3. \(\begin{vmatrix} 4 & 1 \\ 5 & -4 \end{vmatrix} = (4)(-4) - (1)(5) = -16 - 5 = -21\)

Now substituting back:
\[ V = 3(-10) + 1(23) + 3(-21) \]
\[ = -30 + 23 - 63 = -70 \]

### Step 7: Find the absolute value
Since volume is always positive:
\[ V = | -70 | = 70 \]

### Final Answer
The volume of the parallelepiped is \(70\) cubic units.

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