ABCD is a parallelogram with \(\overrightarrow{AB} = \vec{a}\) and \(\overrightarrow{AD} = \vec{b}\). Which of the following statements are correct?
(A) \(\overrightarrow{AC} = \vec{a}+\vec{b}\)
(B) \(\overrightarrow{BD} = \vec{b}-\vec{a}\)
(C) The area of ABCD is \(\displaystyle\frac{1}{2}|\vec{a}\times\vec{b}|\)
(D) If \(|\vec{a}| = |\vec{b}|\), then the diagonals AC and BD are perpendicular
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A), (B) and (D) only
In a parallelogram \(\overrightarrow{BC} = \overrightarrow{AD} = \vec{b}\). By the triangle law:
\(\overrightarrow{AC} = \overrightarrow{AB}+\overrightarrow{BC} = \vec{a}+\vec{b}\)
\(\overrightarrow{BD} = \overrightarrow{AD}-\overrightarrow{AB} = \vec{b}-\vec{a}\)
The scalar product of the diagonals is
\((\vec{a}+\vec{b})\cdot(\vec{b}-\vec{a}) = |\vec{b}|^2-|\vec{a}|^2\)
So \(\overrightarrow{AC} = \vec{a}+\vec{b}\), \(\overrightarrow{BD} = \vec{b}-\vec{a}\), the area is \(|\vec{a}\times\vec{b}|\), and the diagonals are perpendicular exactly when \(|\vec{a}| = |\vec{b}|\).
Checking the statements
(A) \(\overrightarrow{AC} = \vec{a}+\vec{b}\) — Correct. \(\overrightarrow{AB}+\overrightarrow{BC}\) with \(\overrightarrow{BC} = \vec{b}\).
(B) \(\overrightarrow{BD} = \vec{b}-\vec{a}\) — Correct. \(\overrightarrow{BA}+\overrightarrow{AD} = -\vec{a}+\vec{b}\).
(C) Area \(= \frac{1}{2}|\vec{a}\times\vec{b}|\) — Incorrect. The parallelogram area is \(|\vec{a}\times\vec{b}|\); half of it is the area of triangle ABD.
(D) \(|\vec{a}| = |\vec{b}|\) makes AC and BD perpendicular — Correct. \((\vec{a}+\vec{b})\cdot(\vec{b}-\vec{a}) = |\vec{b}|^2-|\vec{a}|^2 = 0\).