If `veca=2hati-3hatj+5hatk , vecb=3hati-4hatj+5hatk` and `vecc=5hati-3hatj-2hatk`, then the volume of the parallelopiped with coterminous edges `veca+vecb,vecb+vecc,vecc+veca` is

Options

A. 2
B. 1
C. -1
D. 0

Explanation

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 can use the scalar triple product, which is given by the determinant of a matrix formed by these vectors.

### Step 1: Calculate the vectors \(\vec{a} + \vec{b}\), \(\vec{b} + \vec{c}\), and \(\vec{c} + \vec{a}\)

Given:
\[ \vec{a} = 2\hat{i} - 3\hat{j} + 5\hat{k} \]
\[ \vec{b} = 3\hat{i} - 4\hat{j} + 5\hat{k} \]
\[ \vec{c} = 5\hat{i} - 3\hat{j} - 2\hat{k} \]

Now, we calculate:
1. \(\vec{a} + \vec{b}\):
\[ \vec{a} + \vec{b} = (2 + 3)\hat{i} + (-3 - 4)\hat{j} + (5 + 5)\hat{k} = 5\hat{i} - 7\hat{j} + 10\hat{k} \]

2. \(\vec{b} + \vec{c}\):
\[ \vec{b} + \vec{c} = (3 + 5)\hat{i} + (-4 - 3)\hat{j} + (5 - 2)\hat{k} = 8\hat{i} - 7\hat{j} + 3\hat{k} \]

3. \(\vec{c} + \vec{a}\):
\[ \vec{c} + \vec{a} = (5 + 2)\hat{i} + (-3 - 3)\hat{j} + (-2 + 5)\hat{k} = 7\hat{i} - 6\hat{j} + 3\hat{k} \]

### Step 2: Set up the determinant

The volume \(V\) of the parallelepiped is given by the absolute value of the scalar triple product:
\[ V = |\vec{u} \cdot (\vec{v} \times \vec{w})| \]
where \(\vec{u} = \vec{a} + \vec{b}\), \(\vec{v} = \vec{b} + \vec{c}\), and \(\vec{w} = \vec{c} + \vec{a}\).

This can be calculated using the determinant:
\[ V = \left| \begin{vmatrix} 5 & -7 & 10 \\ 8 & -7 & 3 \\ 7 & -6 & 3 \end{vmatrix} \right| \]

### Step 3: Calculate the determinant

We will calculate the determinant using cofactor expansion along the first row:
\[ \begin{vmatrix} 5 & -7 & 10 \\ 8 & -7 & 3 \\ 7 & -6 & 3 \end{vmatrix} = 5 \begin{vmatrix} -7 & 3 \\ -6 & 3 \end{vmatrix} - (-7) \begin{vmatrix} 8 & 3 \\ 7 & 3 \end{vmatrix} + 10 \begin{vmatrix} 8 & -7 \\ 7 & -6 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. \(\begin{vmatrix} -7 & 3 \\ -6 & 3 \end{vmatrix} = (-7)(3) - (-6)(3) = -21 + 18 = -3\)

2. \(\begin{vmatrix} 8 & 3 \\ 7 & 3 \end{vmatrix} = (8)(3) - (7)(3) = 24 - 21 = 3\)

3. \(\begin{vmatrix} 8 & -7 \\ 7 & -6 \end{vmatrix} = (8)(-6) - (7)(-7) = -48 + 49 = 1\)

Substituting back into the determinant:
\[ = 5(-3) + 7(3) + 10(1) = -15 + 21 + 10 = 16 \]

### Step 4: Find the volume

Thus, the volume of the parallelepiped is:
\[ V = |16| = 16 \]

### Final Answer:
The volume of the parallelepiped is \(16\).

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