Question

If \( A, B \) and \( C \) be three non-collinear points such that \( \vec{AB} = \hat{i} + 2\hat{j} - \hat{k} \) and \( \vec{AC} = 2\hat{i} - 3\hat{j} \), then find the area of \( \triangle ABC \).

Hint:

Remember the factor of \( 1/2 \). For a parallelogram, the area is just \( |\vec{a} \times \vec{b}| \), but for a triangle, it must be halved.

Answer 1

Concept:
The area of a triangle \(ABC\), whose two adjacent sides are represented by vectors \(\vec{AB}\) and \(\vec{AC}\), is given by: \[ \text{Area of } \triangle ABC = \frac{1}{2}\left|\vec{AB}\times\vec{AC}\right| \]
Step 1: Calculate the cross product \(\vec{AB}\times\vec{AC}\)
Given, \[ \vec{AB}=(1,2,-1) \] and \[ \vec{AC}=(2,-3,0) \] Using the determinant method: \[ \vec{AB}\times\vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -1 \\ 2 & -3 & 0 \end{vmatrix} \] Expanding along the first row: \[ \vec{AB}\times\vec{AC} = \hat{i}(0-3) - \hat{j}(0-(-2)) + \hat{k}(-3-4) \] \[ = -3\hat{i}-2\hat{j}-7\hat{k} \] 
Step 2: Find the magnitude of the cross product
\[ \left|\vec{AB}\times\vec{AC}\right| = \sqrt{(-3)^2+(-2)^2+(-7)^2} \] \[ = \sqrt{9+4+49} = \sqrt{62} \] 
Step 3: Calculate the area of the triangle
Using \[ \text{Area of } \triangle ABC = \frac{1}{2}\left|\vec{AB}\times\vec{AC}\right| \] we get \[ \text{Area of } \triangle ABC = \frac{\sqrt{62}}{2} \] \[ \text{Area} \approx 3.94 \] 
Final Answer:
\[ \boxed{\frac{\sqrt{62}}{2}\text{ square units}} \] Approximately, \[ \boxed{3.94\text{ square units}} \]