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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class12  >>  Vector Algebra
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If $\overrightarrow a,\overrightarrow b\:\overrightarrow c$ are three vectors of equal magnitude and angle between $\overrightarrow a\:and\:\overrightarrow b$ is $\alpha$, that between $\overrightarrow b\:\overrightarrow c$ is $\beta$, and that between $\overrightarrow a\:and\:\overrightarrow c$ is $\gamma$ then the minimum value of $cos\alpha+cos\beta+cos\gamma$ is ?

$\begin{array}{1 1} \frac{1}{2} \\ \frac{-1}{2} \\ \frac{3}{2} \\ - \frac{3}{2} \end{array} $

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Given: $|\overrightarrow a|=|\overrightarrow b|=|\overrightarrow c|=\:\lambda\:\:\:(say)$
$\overrightarrow a.\overrightarrow b=\lambda^2.cos\alpha$
$\overrightarrow b.\overrightarrow c=\lambda^2.cos\beta$
$\overrightarrow c.\overrightarrow a=\lambda^2.cos\gamma$
We know that $|\overrightarrow a+\overrightarrow b+\overrightarrow c|^2\geq 0$
$\Rightarrow\:|\overrightarrow a|^2+|\overrightarrow b|^2+|\overrightarrow c|^2+2(\overrightarrow a.\overrightarrow b+\overrightarrow b.\overrightarrow c+\overrightarrow a.\overrightarrow a)\geq 0$
$\Rightarrow\:3\lambda^2+2\lambda^2(cos\alpha+cos\beta+cos\gamma)\geq 0$
$\Rightarrow\:cos\alpha+cos\beta+cos\gamma\geq - \large\frac{3}{2}$
$\therefore\:$ The minimum value of $ cos\alpha+cos\beta+cos\gamma $ is $-\large\frac{3}{2}$
answered Nov 15, 2013 by rvidyagovindarajan_1

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