The area of the triangle with vertices A(1, 1, 1), B(2, 3, 1) and C(3, 1, 3) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 3 square units
Two sides of the triangle from A are
\(\overrightarrow{AB} = \hat{\imath}+2\hat{\jmath}, \quad \overrightarrow{AC} = 2\hat{\imath}+2\hat{k}\)
The area of the triangle is \(\displaystyle\frac{1}{2}|\overrightarrow{AB}\times\overrightarrow{AC}|\).
\( \displaystyle \overrightarrow{AB}\times\overrightarrow{AC} = \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 1 & 2 & 0 \\ 2 & 0 & 2 \end{vmatrix} \)
\( = \hat{\imath}(4-0)-\hat{\jmath}(2-0)+\hat{k}(0-4)\)
\( = 4\hat{\imath}-2\hat{\jmath}-4\hat{k}\)
\(|\overrightarrow{AB}\times\overrightarrow{AC}| = \sqrt{16+4+16} = 6\)
So the area is
\( \displaystyle \frac{1}{2}\times 6 = 3 \)