In a triangle ABC, D is the mid-point of the side BC. Which of the following statements are correct?
(A) \(\overrightarrow{AB}+\overrightarrow{AC} = 2\,\overrightarrow{AD}\)
(B) \(\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA} = 2\,\overrightarrow{AC}\)
(C) \(\overrightarrow{AB}-\overrightarrow{AC} = \overrightarrow{CB}\)
(D) \(\overrightarrow{BD}+\overrightarrow{DA} = \overrightarrow{AB}\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) and (C) only
Let \(\vec{a}, \vec{b}, \vec{c}\) be the position vectors of A, B, C. By the mid-point formula, D has position vector
\( \displaystyle \vec{d} = \frac{\vec{b}+\vec{c}}{2} \)
\(\overrightarrow{AB}+\overrightarrow{AC} = (\vec{b}-\vec{a})+(\vec{c}-\vec{a}) = 2\vec{d}-2\vec{a} = 2\,\overrightarrow{AD}\)
\(\overrightarrow{AB}-\overrightarrow{AC} = (\vec{b}-\vec{a})-(\vec{c}-\vec{a}) = \vec{b}-\vec{c} = \overrightarrow{CB}\)
So \(\overrightarrow{AB}+\overrightarrow{AC} = 2\,\overrightarrow{AD}\), \(\overrightarrow{AB}-\overrightarrow{AC} = \overrightarrow{CB}\), going round the triangle gives \(\vec{0}\), and \(\overrightarrow{BD}+\overrightarrow{DA} = \overrightarrow{BA}\) by the triangle law.
Checking the statements
(A) \(\overrightarrow{AB}+\overrightarrow{AC} = 2\,\overrightarrow{AD}\) — Correct. The sum is \(\vec{b}+\vec{c}-2\vec{a} = 2(\vec{d}-\vec{a})\).
(B) \(\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA} = 2\,\overrightarrow{AC}\) — Incorrect. Going A to B to C and back to A gives \(\vec{0}\).
(C) \(\overrightarrow{AB}-\overrightarrow{AC} = \overrightarrow{CB}\) — Correct. The difference is \(\vec{b}-\vec{c}\), the vector from C to B.
(D) \(\overrightarrow{BD}+\overrightarrow{DA} = \overrightarrow{AB}\) — Incorrect. B to D and then D to A ends at A, so the sum is \(\overrightarrow{BA} = -\overrightarrow{AB}\).