Kerala KEAM 5 June 2024 Solved Paper

We are given two vectors: $$\overrightarrow{AB} = \hat{i} + 2\hat{j} - 2\hat{k} \quad \text{and} \quad \overrightarrow{AC} = \hat{i} - \hat{j} + \hat{k}.$$The area of triangle $\triangle ABC$ is given by: $$\text{Area} = \frac{1}{2} \left| \overrightarrow{AB} \times \overrightarrow{AC} \right|.$$Thus, we need to compute the cross product $\overrightarrow{AB} \times \overrightarrow{AC}$.Step 1: Compute the cross product: $$\overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -2 \\ 1 & -1 & 1 \end{vmatrix}.$$Using the determinant formula, we get: $$\overrightarrow{AB} \times \overrightarrow{AC} = \hat{i} \begin{vmatrix} 2 & -2 \\ -1 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & -2 \\ 1 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 2 \\ 1 & -1 \end{vmatrix}.$$This simplifies to: $$= \hat{i} \left( 2 \times 1 - (-2) \times (-1) \right) - \hat{j} \left( 1 \times 1 - (-2) \times 1 \right) + \hat{k} \left( 1 \times (-1) - 2 \times 1 \right)$$ $$= \hat{i} \left( 2 - 2 \right) - \hat{j} \left( 1 - (-2) \right) + \hat{k} \left( -1 - 2 \right)$$ $$= 0\hat{i} - 3\hat{j} - 3\hat{k}.$$Thus, $$\overrightarrow{AB} \times \overrightarrow{AC} = -3\hat{j} - 3\hat{k}.$$Step 2: Find the magnitude of the cross product: $$\left| \overrightarrow{AB} \times \overrightarrow{AC} \right| = \sqrt{(-3)^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}.$$Step 3: Finally, the area of triangle $\triangle ABC$ is: $$\text{Area} = \frac{1}{2} \times 3\sqrt{2} = \frac{3}{\sqrt{2}}.$$Thus, the correct answer is option (D).Quick Tip: The area of a triangle formed by two vectors can be found using the formula $\text{Area} = \frac{1}{2} \left| \overrightarrow{AB} \times \overrightarrow{AC} \right|$. Make sure to compute the cross product and find its magnitude to get the correct answer.