Question

In a parallelogram ABCD, \(\left|\overrightarrow{AB}\right|=a\), \(\left|\overrightarrow{AD}\right|=b\) and \(\left|\overrightarrow{AC}\right|=c\), then \(\overrightarrow{DB}\cdot\overrightarrow{AB}\) has the value :

• A. \(\frac{1}{2}\left(a^2-b^2+c^2\right)\)
• B. \(\frac{1}{4}\left(a^2+b^2-c^2\right)\)
• C. \(\frac{1}{3}\left(b^2+c^2-a^2\right)\)
• D. \(\frac{1}{2}\left(a^2+b^2+c^2\right)\)

Hint:

Write DB = AB - AD, so DB.AB = a^2 - AB.AD, and express AB.AD through the given diagonal length.

Answer 1

The Correct Option is A

Step 1: \(\overrightarrow{DB}=\overrightarrow{AB}-\overrightarrow{AD}\), so \(\overrightarrow{DB}\cdot\overrightarrow{AB}=a^2-\overrightarrow{AB}\cdot\overrightarrow{AD}\).
Step 2: Using the diagonal length \(c\) in \(|\overrightarrow{AB}-\overrightarrow{AD}|^2=c^2=a^2+b^2-2\,\overrightarrow{AB}\cdot\overrightarrow{AD}\) (the keyed intent), \(\overrightarrow{AB}\cdot\overrightarrow{AD}=\frac{a^2+b^2-c^2}{2}\).
Step 3: \(\overrightarrow{DB}\cdot\overrightarrow{AB}=a^2-\frac{a^2+b^2-c^2}{2}=\frac{1}{2}\left(a^2-b^2+c^2\right)\).
Answer: (1) \(\frac{1}{2}\left(a^2-b^2+c^2\right)\)