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 \(\Delta ABC\).

Hint:

Always check the order of components in the determinant.
The cross product method is significantly faster for triangles in 3D space compared to coordinates-based formulas.

Answer 1

Concept:
The area of a triangle with adjacent sides \(\vec{AB}\) and \(\vec{AC}\) is given by: \[ \text{Area}=\frac{1}{2}|\vec{AB}\times\vec{AC}| \] The cross product can be calculated using the determinant method. 
Step 1: Calculate \(\vec{AB}\times\vec{AC}\)
Given: \[ \vec{AB}=(1,2,-1), \qquad \vec{AC}=(2,-3,0) \] Therefore, \[ \vec{AB}\times\vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -1 \\ 2 & -3 & 0 \end{vmatrix} \] Expanding the determinant: \[ =\hat{i}(2\cdot0-(-1)(-3)) -\hat{j}(1\cdot0-(-1)(2)) +\hat{k}(1(-3)-2(2)) \] \[ =-3\hat{i}-2\hat{j}-7\hat{k} \] 
Step 2: Find the magnitude of the cross product
\[ |\vec{AB}\times\vec{AC}| = \sqrt{(-3)^2+(-2)^2+(-7)^2} \] \[ =\sqrt{9+4+49} \] \[ =\sqrt{62} \] 
Step 3: Calculate the area of the triangle
Using the area formula: \[ \text{Area of }\triangle ABC = \frac{1}{2}|\vec{AB}\times\vec{AC}| \] \[ =\frac{\sqrt{62}}{2} \] 
Final Answer:
Therefore, the area of the triangle is: \[ \boxed{\frac{\sqrt{62}}{2}\text{ sq. units}} \]