Question

If $A$, $B$ and $C$ are three non-collinear points in a plane such that their relative position vectors form the directed sides $\vec{AB} = \hat{i} + 2\hat{j} - \hat{k}$ and $\vec{AC} = 2\hat{i} - 3\hat{j}$, then compute the exact geometric area of the triangle $\Delta ABC$.

Hint:

When expanding the cross product determinant, remember that the middle term ($\hat{j}$) always carries a negative sign by default from matrix cofactor rules: $+\hat{i}, -\hat{j}, +\hat{k}$.

Answer 1

Concept: In vector geometry, the area of a triangle formed by three vertices $A$, $B$, and $C$ can be evaluated using the cross product of any two adjacent directed vectors originating from a shared vertex. If we select vertex $A$ as our base origin point, the two adjacent vector sides are $\vec{AB}$ and $\vec{AC}$. The mathematical formula for the area of triangle $\Delta ABC$ is given by: \[ \text{Area} = \frac{1}{2} \left| \vec{AB} \times \vec{AC} \right| \] where $\vec{AB} \times \vec{AC}$ denotes the vector cross product, and $\left| \cdot \right| $ represents the spatial magnitude of the resulting cross product vector.

Step 1: Formulating and computing the vector cross product $\vec{AB} \times \vec{AC}$ using a matrix determinant.
Let us express the vectors with all three Cartesian components explicitly visible: \[ \vec{AB} = 1\hat{i} + 2\hat{j} - 1\hat{k} \] \[ \vec{AC} = 2\hat{i} - 3\hat{j} + 0\hat{k} \] The cross product is computed by expanding a $3 \times 3$ matrix determinant layout: \[ \vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -1 2 & -3 & 0 \end{vmatrix} \] Expanding this determinant along the first row: \[ \vec{AB} \times \vec{AC} = \hat{i} \begin{vmatrix} 2 & -1 -3 & 0 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & -1 \\ 2 & 0 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 2\\ 2 & -3 \end{vmatrix} \] Now, compute the values of individual $2 \times 2$ determinants:
• For $\hat{i}$: $(2)(0) - (-1)(-3) = 0 - 3 = -3$
• For $-\hat{j}$: $(1)(0) - (-1)(2) = 0 - (-2) = 2$
• For $\hat{k}$: $(1)(-3) - (2)(2) = -3 - 4 = -7$ Combining these components gives: \[ \vec{AB} \times \vec{AC} = -3\hat{i} - 2\hat{j} - 7\hat{k} \]

Step 2: Calculating the absolute spatial magnitude of the cross product vector.
The magnitude of our cross product vector is determined using the standard three-dimensional Pythagorean distance formula: \[ \left| \vec{AB} \times \vec{AC} \right| = \sqrt{(-3)^2 + (-2)^2 + (-7)^2} \] Squaring each real coefficient component: \[ \left| \vec{AB} \times \vec{AC} \right| = \sqrt{9 + 4 + 49} \] Summing the values under the square root: \[ \left| \vec{AB} \times \vec{AC} \right| = \sqrt{62} \]

Step 3: Substituting the magnitude value into the triangle area formula.
Now, apply the triangle area formula: \[ \text{Area} = \frac{1}{2} \left| \vec{AB} \times \vec{AC} \right| = \frac{1}{2} \cdot \sqrt{62} = \frac{\sqrt{62}}{2}\text{ square units} \] Thus, the final exact area of the triangle is $\frac{\sqrt{62}}{2}$ square units.