A, B and C are the points (2, 1, 3), (3, 2, 3) and (1, 1, 4) respectively. The angle ABC of the triangle ABC is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{\pi}{6}\)
The angle ABC is the angle between the two sides that start at B, that is, between \(\overrightarrow{BA}\) and \(\overrightarrow{BC}\):
\(\overrightarrow{BA} = (2-3)\hat{\imath}+(1-2)\hat{\jmath}+(3-3)\hat{k} = -\hat{\imath}-\hat{\jmath}\)
\(\overrightarrow{BC} = (1-3)\hat{\imath}+(1-2)\hat{\jmath}+(4-3)\hat{k} = -2\hat{\imath}-\hat{\jmath}+\hat{k}\)
\(\overrightarrow{BA}\cdot\overrightarrow{BC} = 2+1+0 = 3\)
\(|\overrightarrow{BA}| = \sqrt{2}, \quad |\overrightarrow{BC}| = \sqrt{6}\)
\( \displaystyle \cos B = \frac{3}{\sqrt{2}\,\sqrt{6}} = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2} \)
\( \displaystyle B = \frac{\pi}{6} \)
Why the other options are incorrect
Option (2) \(\displaystyle\frac{\pi}{3}\) — Incorrect. \(\cos\displaystyle\frac{\pi}{3} = \frac{1}{2}\), but \(\cos B = \displaystyle\frac{\sqrt{3}}{2}\).
Option (3) \(\displaystyle\frac{2\pi}{3}\) — Incorrect. This is the angle at A: \(\overrightarrow{AB}\cdot\overrightarrow{AC} = (\hat{\imath}+\hat{\jmath})\cdot(-\hat{\imath}+\hat{k}) = -1\), giving \(\cos A = -\displaystyle\frac{1}{2}\).
Option (4) \(\displaystyle\frac{5\pi}{6}\) — Incorrect. This comes from using \(\overrightarrow{AB}\) with \(\overrightarrow{BC}\); one of these points into B, so their angle is \(\pi-B\), not B.