Physics · 2019
JEE Main · 8 April 2019, Shift 2 · Q2
Let | A_1 |=3,| A_2 |=5 and | A_1 + A_2 |=5. The value of (2 A_1 +3 A_2 ) ·(3 A_1 -2 A_2 ) is:
Let $\displaystyle \left|\overrightarrow{\mathrm{A}_1}\right|=3,\left|\overrightarrow{\mathrm{~A}_2}\right|=5$ and $\displaystyle \left|\overrightarrow{\mathrm{A}_1}+\overrightarrow{\mathrm{A}_2}\right|=5$. The
value of $\displaystyle \left(2 \overrightarrow{\mathrm{~A}_1}+3 \overrightarrow{\mathrm{~A}_2}\right) \cdot\left(3 \overrightarrow{\mathrm{~A}_1}-2 \overrightarrow{\mathrm{~A}_2}\right)$ is:
Official answer
(4)
$\displaystyle -118.5$
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JEE Main 2019 Physics question, with the answer from NTA’s final answer key. Where our answers come from.