Area (in sq. units) of the triangle with vertices A (1,1,1) B (1,2,3) and C (2,3,1) is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \displaystyle\frac{1}{2}\sqrt{21} \)
The area of triangle ABC is \( \frac{1}{2}\left|\overrightarrow{AB}\times\overrightarrow{AC}\right| \).
\( \displaystyle \overrightarrow{AB}=0\hat{\imath}+\hat{\jmath}+2\hat{k} \)
\( \displaystyle \overrightarrow{AC}=\hat{\imath}+2\hat{\jmath}+0\hat{k} \)
\( \displaystyle \overrightarrow{AB}\times\overrightarrow{AC}=\begin{vmatrix}\hat{\imath} & \hat{\jmath} & \hat{k}\\0 & 1 & 2\\1 & 2 & 0\end{vmatrix} \)
\( \displaystyle =\hat{\imath}(0-4)-\hat{\jmath}(0-2)+\hat{k}(0-1) \)
\( \displaystyle =-4\hat{\imath}+2\hat{\jmath}-\hat{k} \)
\( \displaystyle \left|\overrightarrow{AB}\times\overrightarrow{AC}\right|=\sqrt{16+4+1}=\sqrt{21} \)
\( \displaystyle \text{Area}=\frac{1}{2}\sqrt{21}\ \text{sq. units} \)
Why the other options are incorrect
Option (2) \( \sqrt{21} \) — Incorrect. \( \sqrt{21} \) is \( \left|\overrightarrow{AB}\times\overrightarrow{AC}\right| \), the area of the parallelogram on AB and AC; the triangle is half of it.
Option (3) 3 — Incorrect. The area is \( \frac{\sqrt{21}}{2}\approx 2.29 \), not 3.
Option (4) \( \displaystyle\frac{3}{2} \) — Incorrect. The area is \( \frac{\sqrt{21}}{2}\approx 2.29 \), not 1.5.
NCERT (2026-27 edition):
Product of Two Vectors
