Let `overset(to)(a_(r))=x_(r )hat(i)+y_(r )hat(j)+z_(r )hat(k),r=1,2,3` three mutually prependicular unit vectors then the value of `|{:(x_(1),,-x_(2),,x_(3)),(y_(1),,y_(2) ,,y_(3)),(z_(1),,z_(2),,z_(3)):}|` is equal to

Options

A. 0
B. `pm1`
C. `pm2`
D. `pm4`

Answer

To solve the problem, we need to find the value of the determinant given by:

\[ D = \begin{vmatrix} x_1 & -x_2 & x_3 \\ y_1 & y_2 & y_3 \\ z_1 & z_2 & z_3 \end{vmatrix} \]

where \( \overset{\to}{a_r} = x_r \hat{i} + y_r \hat{j} + z_r \hat{k} \) for \( r = 1, 2, 3 \) represents three mutually perpendicular unit vectors.

### Step 1: Write the determinant

We can express the determinant \( D \) as follows:

\[ D = \begin{vmatrix} x_1 & -x_2 & x_3 \\ y_1 & y_2 & y_3 \\ z_1 & z_2 & z_3 \end{vmatrix} \]

### Step 2: Expand the determinant

Using the determinant expansion formula, we can expand \( D \):

\[ D = x_1 \begin{vmatrix} y_2 & y_3 \\ z_2 & z_3 \end{vmatrix} - (-x_2) \begin{vmatrix} y_1 & y_3 \\ z_1 & z_3 \end{vmatrix} + x_3 \begin{vmatrix} y_1 & y_2 \\ z_1 & z_2 \end{vmatrix} \]

### Step 3: Calculate the 2x2 determinants

Calculating the 2x2 determinants:

1. \( \begin{vmatrix} y_2 & y_3 \\ z_2 & z_3 \end{vmatrix} = y_2 z_3 - y_3 z_2 \)

2. \( \begin{vmatrix} y_1 & y_3 \\ z_1 & z_3 \end{vmatrix} = y_1 z_3 - y_3 z_1 \)

3. \( \begin{vmatrix} y_1 & y_2 \\ z_1 & z_2 \end{vmatrix} = y_1 z_2 - y_2 z_1 \)

### Step 4: Substitute back into the determinant

Substituting these back into our expression for \( D \):

\[ D = x_1 (y_2 z_3 - y_3 z_2) + x_2 (y_1 z_3 - y_3 z_1) + x_3 (y_1 z_2 - y_2 z_1) \]

### Step 5: Recognize the structure

Notice that the expression for \( D \) is the scalar triple product of the vectors \( \overset{\to}{a_1}, \overset{\to}{a_2}, \overset{\to}{a_3} \). Since these vectors are mutually perpendicular unit vectors, the scalar triple product equals the volume of the parallelepiped formed by these vectors, which is 1 (since the volume of a unit cube is 1).

### Step 6: Conclusion

Thus, the value of the determinant \( D \) is:

\[ D = 1 \]

### Final Answer

The value of the determinant is \( \pm 1 \).

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