A, B and C are three distinct points and O is any point. If \(\overrightarrow{AO}+\overrightarrow{OB} = \overrightarrow{BO}+\overrightarrow{OC}\), then A, B and C are
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → collinear, with B the mid-point of AC
By the triangle law, going from A to O and then from O to B is the same as going from A to B:
\( \overrightarrow{AO}+\overrightarrow{OB} = \overrightarrow{AB}, \quad \overrightarrow{BO}+\overrightarrow{OC} = \overrightarrow{BC} \)
\( \overrightarrow{AB} = \overrightarrow{BC} \)
With \(\vec{a}, \vec{b}, \vec{c}\) the position vectors of A, B, C:
\( \vec{b}-\vec{a} = \vec{c}-\vec{b} \)
\( \displaystyle \vec{b} = \frac{\vec{a}+\vec{c}}{2} \)
So B is the mid-point of AC, and the three points lie on one line.
Why the other options are incorrect
Option (1) the vertices of an equilateral triangle — Incorrect. Since \(\overrightarrow{AB}\) and \(\overrightarrow{BC}\) are equal vectors, the three points lie on one line and form no triangle.
Option (2) the vertices of a right-angled triangle — Incorrect. The points are collinear, so there is no triangle and no right angle.
Option (3) collinear, with A the mid-point of BC — Incorrect. A would be the mid-point of BC if \(\overrightarrow{BA} = \overrightarrow{AC}\); here the equal vectors are \(\overrightarrow{AB}\) and \(\overrightarrow{BC}\), which meet at B.