The value ( in cubic units ) of the parallelopiped whose coterminus edges are given by the vectors `vec(i)-vec(j)+hat(k),2hat(i)-4hat(j)+5hat(k)` and `3 hat(i) -5 hat(j) + 2hat(k)` is

Options

A. 2
B. 3
C. 4
D. 8

Answer

To find the volume of the parallelepiped formed by the vectors \(\vec{a} = \hat{i} - \hat{j} + \hat{k}\), \(\vec{b} = 2\hat{i} - 4\hat{j} + 5\hat{k}\), and \(\vec{c} = 3\hat{i} - 5\hat{j} + 2\hat{k}\), we will calculate the scalar triple product, which can be represented as the determinant of a 3x3 matrix formed by these vectors.

### Step 1: Write the vectors in matrix form
We can represent the vectors as follows:

\[ \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 2 & -4 & 5 \\ 3 & -5 & 2 \end{vmatrix} \]

### Step 2: Calculate the determinant
The volume \(V\) of the parallelepiped is given by the absolute value of the determinant of the matrix formed by the vectors:

\[ V = \left| \begin{vmatrix} 1 & -1 & 1 \\ 2 & -4 & 5 \\ 3 & -5 & 2 \end{vmatrix} \right| \]

To calculate this determinant, we can use the formula for the determinant of a 3x3 matrix:

\[ \text{det} = a(ei - fh) - b(di - fg) + c(dh - eg) \]

Where the matrix is:

\[ \begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} \]

In our case:

- \(a = 1, b = -1, c = 1\)
- \(d = 2, e = -4, f = 5\)
- \(g = 3, h = -5, i = 2\)

Calculating the determinant:

\[ \text{det} = 1((-4)(2) - (5)(-5)) - (-1)((2)(2) - (5)(3)) + 1((2)(-5) - (-4)(3)) \]

Calculating each term step by step:

1. First term:
\[ 1((-4)(2) - (5)(-5)) = 1(-8 + 25) = 1(17) = 17 \]

2. Second term:
\[ -(-1)((2)(2) - (5)(3)) = 1(4 - 15) = 1(-11) = -11 \]

3. Third term:
\[ 1((2)(-5) - (-4)(3)) = 1(-10 + 12) = 1(2) = 2 \]

Now, summing these results:

\[ \text{det} = 17 - 11 + 2 = 8 \]

### Step 3: Conclusion
Thus, the volume of the parallelepiped is:

\[ V = |8| = 8 \text{ cubic units} \]