Question

Let \( \overrightarrow{AB} = i + 2j - 2k \) and \( \overrightarrow{AC} = i - j + k. \) Then the area of \( \triangle ABC \) is:

• A. \( 3\sqrt{2} \)
• B. \( \frac{3}{2} \)
• C. \( 2\sqrt{3} \)
• D. \( \frac{3}{\sqrt{2}} \)
• E. \( \frac{1}{\sqrt{2}} \)

Hint:

The area of a triangle formed by two vectors can be found using the formula \( {Area} = \frac{1}{2} \left| \overrightarrow{AB} \times \overrightarrow{AC} \right| \). Make sure to compute the cross product and find its magnitude to get the correct answer.

Answer 1

The Correct Option is D

We are given two vectors:

\[ \overrightarrow{AB} = i + 2j - 2k \quad \text{and} \quad \overrightarrow{AC} = i - j + k. \] 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).