Question

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

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

Hint:

You can use any two vectors starting from the same vertex (e.g., \(\vec{BA}\) and \(\vec{BC}\)) to get the same result.
Double-check determinant expansion signs: \(+ \hat{i}, - \hat{j}, + \hat{k}\).

Answer 1

The Correct Option is A

Concept:
The area of a triangle with vertices \(A\), \(B\), and \(C\) is given by: \[ \text{Area}=\frac{1}{2}\left|\vec{AB}\times\vec{AC}\right| \] The vectors are obtained using: \[ \vec{AB}=\vec{B}-\vec{A}, \qquad \vec{AC}=\vec{C}-\vec{A} \] 
Step 1: Find \(\vec{AB}\) and \(\vec{AC}\)
\[ \vec{AB} =(2-0)\hat{i}+(0-1)\hat{j}+(-1-1)\hat{k} \] \[ \vec{AB}=2\hat{i}-\hat{j}-2\hat{k} \] Similarly: \[ \vec{AC} =(1-0)\hat{i}+(0-1)\hat{j}+(3-1)\hat{k} \] \[ \vec{AC}=\hat{i}-\hat{j}+2\hat{k} \] 
Step 2: Calculate \(\vec{AB}\times\vec{AC}\)
\[ \vec{AB}\times\vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & -2 \\ 1 & -1 & 2 \end{vmatrix} \] \[ =\hat{i}[(-1)(2)-(-2)(-1)] -\hat{j}[(2)(2)-(-2)(1)] +\hat{k}[(2)(-1)-(-1)(1)] \] \[ =\hat{i}(-2-2)-\hat{j}(4+2)+\hat{k}(-2+1) \] \[ \vec{AB}\times\vec{AC} =-4\hat{i}-6\hat{j}-\hat{k} \] 
Step 3: Find the magnitude of the cross product
\[ |\vec{AB}\times\vec{AC}| = \sqrt{(-4)^2+(-6)^2+(-1)^2} \] \[ =\sqrt{16+36+1} \] \[ =\sqrt{53} \] 
Step 4: Calculate the area of the triangle
\[ \text{Area} = \frac{1}{2}|\vec{AB}\times\vec{AC}| \] \[ \text{Area} = \frac{1}{2}\sqrt{53} \] 
Final Answer:
Therefore, the area of the triangle is: \[ \boxed{\frac{\sqrt{53}}{2}\text{ sq. units}} \]