Question

The area of the triangle whose vertices are $A(1,-1,2)$, $B(2,1,-1)$ and $C(3,-1,2)$ is

• A. $\sqrt{7}$
• B. $\sqrt{11}$
• C. $\sqrt{13}$
• D. $\sqrt{15}$
• E. $\sqrt{10}$

Hint:

Cross product magnitude directly gives twice the triangle area.

Answer 1

The Correct Option is C

Concept: Area of triangle: \[ \frac{1}{2} |\vec{AB} \times \vec{AC}| \]

Step 1: Compute vectors.
\[ \vec{AB} = (2-1, 1+1, -1-2) = (1,2,-3) \] \[ \vec{AC} = (3-1, -1+1, 2-2) = (2,0,0) \]

Step 2: Cross product.
\[ \vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -3 \\ 2 & 0 & 0 \end{vmatrix} \] \[ = \hat{i}(2\cdot0 - (-3)\cdot0) - \hat{j}(1\cdot0 - (-3)\cdot2) + \hat{k}(1\cdot0 - 2\cdot2) \] \[ = (0, -6, -4) \]

Step 3: Magnitude.
\[ |\vec{AB} \times \vec{AC}| = \sqrt{0 + 36 + 16} = \sqrt{52} = 2\sqrt{13} \]

Step 4: Area.
\[ \frac{1}{2} \times 2\sqrt{13} = \sqrt{13} \] \[ \boxed{\sqrt{13}} \]