Answer :
To find the area of the triangle formed by the vectors [tex]\(\mathbf{A} = \mathbf{i} + \mathbf{j} + \mathbf{k}\)[/tex] and [tex]\(\mathbf{B} = 2\mathbf{i} - \mathbf{j} + 3\mathbf{k}\)[/tex], we follow these steps:
1. Represent the Vectors in Component Form:
[tex]\[ \mathbf{A} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}, \quad \mathbf{B} = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix} \][/tex]
2. Calculate the Cross Product:
The cross product [tex]\(\mathbf{A} \times \mathbf{B}\)[/tex] of two vectors [tex]\(\mathbf{A} = \begin{pmatrix} A_1 \\ A_2 \\ A_3 \end{pmatrix}\)[/tex] and [tex]\(\mathbf{B} = \begin{pmatrix} B_1 \\ B_2 \\ B_3 \end{pmatrix}\)[/tex] is given by:
[tex]\[ \mathbf{A} \times \mathbf{B} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 1 & 1 \\ 2 & -1 & 3 \end{vmatrix} = \mathbf{i}(1 \cdot 3 - 1 \cdot (-1)) - \mathbf{j}(1 \cdot 3 - 1 \cdot 2) + \mathbf{k}(1 \cdot (-1) - 1 \cdot 2) \][/tex]
Simplifying the determinants:
[tex]\[ \mathbf{A} \times \mathbf{B} = \mathbf{i}(3 + 1) - \mathbf{j}(3 - 2) + \mathbf{k}(-1 - 2) = 4\mathbf{i} - \mathbf{j} - 3\mathbf{k} \][/tex]
Thus, the cross product is:
[tex]\[ \mathbf{A} \times \mathbf{B} = \begin{pmatrix} 4 \\ -1 \\ -3 \end{pmatrix} \][/tex]
3. Find the Magnitude of the Cross Product:
The magnitude of the vector [tex]\(\begin{pmatrix} x \\ y \\ z \end{pmatrix}\)[/tex] is given by:
[tex]\[ \|\begin{pmatrix} 4 \\ -1 \\ -3 \end{pmatrix}\| = \sqrt{4^2 + (-1)^2 + (-3)^2} = \sqrt{16 + 1 + 9} = \sqrt{26} \][/tex]
4. Calculate the Area of the Triangle:
The area of the triangle formed by vectors [tex]\(\mathbf{A}\)[/tex] and [tex]\(\mathbf{B}\)[/tex] is half the magnitude of their cross product:
[tex]\[ \text{Area} = \frac{1}{2} \times \|\mathbf{A} \times \mathbf{B}\| = \frac{1}{2} \times \sqrt{26} = \frac{\sqrt{26}}{2} \, \text{square units} \][/tex]
Thus, the area of the triangle is:
C. [tex]\(\frac{\sqrt{26}}{2}\)[/tex] Square units
1. Represent the Vectors in Component Form:
[tex]\[ \mathbf{A} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}, \quad \mathbf{B} = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix} \][/tex]
2. Calculate the Cross Product:
The cross product [tex]\(\mathbf{A} \times \mathbf{B}\)[/tex] of two vectors [tex]\(\mathbf{A} = \begin{pmatrix} A_1 \\ A_2 \\ A_3 \end{pmatrix}\)[/tex] and [tex]\(\mathbf{B} = \begin{pmatrix} B_1 \\ B_2 \\ B_3 \end{pmatrix}\)[/tex] is given by:
[tex]\[ \mathbf{A} \times \mathbf{B} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 1 & 1 \\ 2 & -1 & 3 \end{vmatrix} = \mathbf{i}(1 \cdot 3 - 1 \cdot (-1)) - \mathbf{j}(1 \cdot 3 - 1 \cdot 2) + \mathbf{k}(1 \cdot (-1) - 1 \cdot 2) \][/tex]
Simplifying the determinants:
[tex]\[ \mathbf{A} \times \mathbf{B} = \mathbf{i}(3 + 1) - \mathbf{j}(3 - 2) + \mathbf{k}(-1 - 2) = 4\mathbf{i} - \mathbf{j} - 3\mathbf{k} \][/tex]
Thus, the cross product is:
[tex]\[ \mathbf{A} \times \mathbf{B} = \begin{pmatrix} 4 \\ -1 \\ -3 \end{pmatrix} \][/tex]
3. Find the Magnitude of the Cross Product:
The magnitude of the vector [tex]\(\begin{pmatrix} x \\ y \\ z \end{pmatrix}\)[/tex] is given by:
[tex]\[ \|\begin{pmatrix} 4 \\ -1 \\ -3 \end{pmatrix}\| = \sqrt{4^2 + (-1)^2 + (-3)^2} = \sqrt{16 + 1 + 9} = \sqrt{26} \][/tex]
4. Calculate the Area of the Triangle:
The area of the triangle formed by vectors [tex]\(\mathbf{A}\)[/tex] and [tex]\(\mathbf{B}\)[/tex] is half the magnitude of their cross product:
[tex]\[ \text{Area} = \frac{1}{2} \times \|\mathbf{A} \times \mathbf{B}\| = \frac{1}{2} \times \sqrt{26} = \frac{\sqrt{26}}{2} \, \text{square units} \][/tex]
Thus, the area of the triangle is:
C. [tex]\(\frac{\sqrt{26}}{2}\)[/tex] Square units
