Question

Find the area of the triangle whose two sides are represented by the vectors \[ \vec{a} = 3\hat{i} - \hat{j} + 5\hat{k}, \quad \vec{b} = \hat{i} + 2\hat{j} - \hat{k}. \]

Hint:

The area of a triangle formed by two vectors can be calculated using the cross product formula.

Answer 1

Step 1: Formula for the Area of a Triangle.
The area \( A \) of a triangle formed by two vectors \( \vec{a} \) and \( \vec{b} \) is given by: \[ A = \frac{1}{2} |\vec{a} \times \vec{b}|. \] We need to compute the cross product \( \vec{a} \times \vec{b} \).
Step 2: Compute the Cross Product.
The cross product of \( \vec{a} = 3\hat{i} - \hat{j} + 5\hat{k} \) and \( \vec{b} = \hat{i} + 2\hat{j} - \hat{k} \) is calculated as: \[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -1 & 5 \\ 1 & 2 & -1 \end{vmatrix}. \] Expanding the determinant: \[ \vec{a} \times \vec{b} = \hat{i} \begin{vmatrix} -1 & 5
2 & -1 \end{vmatrix} - \hat{j} \begin{vmatrix} 3 & 5
1 & -1 \end{vmatrix} + \hat{k} \begin{vmatrix} 3 & -1
1 & 2 \end{vmatrix}. \] \[ \vec{a} \times \vec{b} = \hat{i} \left( (-1)(-1) - (5)(2) \right) - \hat{j} \left( (3)(-1) - (5)(1) \right) + \hat{k} \left( (3)(2) - (-1)(1) \right). \] \[ \vec{a} \times \vec{b} = \hat{i} \left( 1 - 10 \right) - \hat{j} \left( -3 - 5 \right) + \hat{k} \left( 6 + 1 \right). \] \[ \vec{a} \times \vec{b} = \hat{i} (-9) - \hat{j} (-8) + \hat{k} (7). \] \[ \vec{a} \times \vec{b} = -9\hat{i} + 8\hat{j} + 7\hat{k}. \]
Step 3: Compute the Magnitude of the Cross Product.
Now, compute the magnitude of \( \vec{a} \times \vec{b} \): \[ |\vec{a} \times \vec{b}| = \sqrt{(-9)^2 + 8^2 + 7^2} = \sqrt{81 + 64 + 49} = \sqrt{194}. \]
Step 4: Compute the Area.
The area of the triangle is: \[ A = \frac{1}{2} \times \sqrt{194}. \]
Step 5: Conclusion.
Thus, the area of the triangle is: \[ A = \frac{\sqrt{194}}{2}. \]