Question

Three points \( A(0, 1, 1) \), \( B(2, 0, -1) \) and \( C(1, 0, 3) \) form \(\Delta ABC\). The area of \(\Delta ABC\) is:

• A. \(\frac{53}{2}\) sq. units
• B. \(\sqrt{53}\) sq. units
• C. \(\frac{\sqrt{53}}{2}\) sq. units
• D. \(\sqrt{11}\) sq. units

Hint:

You can use any pair of vectors pointing from a single vertex, such as \( \vec{BA} \times \vec{BC} \) or \( \vec{CA} \times \vec{CB} \). The resulting cross product vector will always have the exact same magnitude \(\sqrt{53}\).

Answer 1

The Correct Option is C

Concept:
The area of a triangle formed by three points \(A\), \(B\), and \(C\) in 3D space is given by: \[ \text{Area}(\triangle ABC)=\frac{1}{2}\left|\vec{AB}\times\vec{AC}\right| \] where \[ \vec{PQ}=(x_2-x_1)\hat{i}+(y_2-y_1)\hat{j}+(z_2-z_1)\hat{k}. \] 

Step 1: Find \(\vec{AB}\).
Given: \[ A(0,1,1), \qquad B(2,0,-1) \] Therefore, \[ \vec{AB} =(2-0)\hat{i}+(0-1)\hat{j}+(-1-1)\hat{k} =2\hat{i}-\hat{j}-2\hat{k}. \] 

Step 2: Find \(\vec{AC}\).
Given: \[ C(1,0,3) \] Therefore, \[ \vec{AC} =(1-0)\hat{i}+(0-1)\hat{j}+(3-1)\hat{k} =\hat{i}-\hat{j}+2\hat{k}. \] 

Step 3: Find \(\vec{AB}\times\vec{AC}\).
\[ \vec{AB}\times\vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 2 & -1 & -2\\ 1 & -1 & 2 \end{vmatrix} \] Expanding the determinant, \[ \begin{aligned} \vec{AB}\times\vec{AC} &= \hat{i}\left[(-1)(2)-(-2)(-1)\right] -\hat{j}\left[(2)(2)-(-2)(1)\right] +\hat{k}\left[(2)(-1)-(-1)(1)\right] \\ &= \hat{i}(-4)-\hat{j}(6)+\hat{k}(-1) \\ &= -4\hat{i}-6\hat{j}-\hat{k}. \end{aligned} \] 

Step 4: Find the magnitude of the cross product.
\[ \left|\vec{AB}\times\vec{AC}\right| = \sqrt{(-4)^2+(-6)^2+(-1)^2} = \sqrt{16+36+1} = \sqrt{53}. \] 

Step 5: Find the area of the triangle.
\[ \text{Area}(\triangle ABC) = \frac{1}{2}\left|\vec{AB}\times\vec{AC}\right| = \frac{\sqrt{53}}{2} \text{ sq. units}. \] Hence, the correct answer is \[ \boxed{\frac{\sqrt{53}}{2}} \] (Option C).