In a triangle ABC, the point D on the side BC is such that BD : DC = 2 : 1, and E is the mid-point of AD. Then \(\overrightarrow{BE}\) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \displaystyle \frac{1}{6}\left(2\,\overrightarrow{AC}-5\,\overrightarrow{AB}\right) \)
Take A as the origin, so that the position vectors of B and C are \(\vec{b} = \overrightarrow{AB}\) and \(\vec{c} = \overrightarrow{AC}\).
D divides BC internally in the ratio 2 : 1, so by the section formula
\( \displaystyle \vec{d} = \frac{2\vec{c}+1\cdot\vec{b}}{2+1} = \frac{\vec{b}+2\vec{c}}{3} \)
E is the mid-point of AD, and A is the origin:
\( \displaystyle \vec{e} = \frac{\vec{d}}{2} = \frac{\vec{b}+2\vec{c}}{6} \)
\( \displaystyle \overrightarrow{BE} = \vec{e}-\vec{b} = \frac{\vec{b}+2\vec{c}-6\vec{b}}{6} \)
\( \displaystyle \overrightarrow{BE} = \frac{1}{6}\left(2\,\overrightarrow{AC}-5\,\overrightarrow{AB}\right) \)
Why the other options are incorrect
Option (1) \( \displaystyle \frac{1}{6}\left(\overrightarrow{AB}+2\,\overrightarrow{AC}\right) \) — Incorrect. This is \(\overrightarrow{AE}\), the vector from A to E, not from B to E.
Option (2) \( \displaystyle \frac{1}{6}\left(\overrightarrow{AC}-4\,\overrightarrow{AB}\right) \) — Incorrect. This comes from the reversed ratio, \(\vec{d} = \frac{2\vec{b}+\vec{c}}{3}\), which puts D nearer to B.
Option (4) \( \displaystyle \frac{1}{6}\left(5\,\overrightarrow{AB}-2\,\overrightarrow{AC}\right) \) — Incorrect. This is \(\overrightarrow{EB}\), the negative of \(\overrightarrow{BE}\).