Aspire Faculty ID #16569 · Topic: NIMCET 2008 · Just now
NIMCET 2008

If $\vec A+\vec B+\vec C=\vec0$, $|\vec A|=3$, $|\vec B|=5$, $|\vec C|=7$, then the angle between $\vec A$ and $\vec B$ is

Solution

$\vec C=-(\vec A+\vec B)$ $|\vec C|^2=|\vec A|^2+|\vec B|^2+2\vec A\cdot\vec B$ $49=9+25+2(3)(5)\cos\theta$ $49=34+30\cos\theta$ $\cos\theta=\dfrac12 \Rightarrow \theta=\dfrac{\pi}{3}$

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