Mathematics · 2015
JEE Main · 11 April 2015, Shift 1 · Q85
In a parallelogram ABCD,| AB |= a, | AD |= b and | AC |= c, then DB · AB has the value:
In a parallelogram $\displaystyle \mathrm{ABCD},|\overrightarrow{\mathrm{AB}}|=\mathrm{a}$, $\displaystyle |\overrightarrow{\mathrm{AD}}|=\mathrm{b}$ and $\displaystyle |\overrightarrow{\mathrm{AC}}|=\mathrm{c}$, then $\displaystyle \overrightarrow{\mathrm{DB}} \cdot \overrightarrow{\mathrm{AB}}$ has the value :
Official answer
(1)
$\displaystyle \frac{1}{2}\left(a^2-b^2+c^2\right)$
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JEE Main 2015 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.