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Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class12  >>  Vector Algebra

If $\overrightarrow v=2\hat i+\hat j-\hat k, \:\:and\:\:\overrightarrow w=\hat i+3\hat k\:\:and\:\:\overrightarrow u$ is a unit vector, then the maximum value of $[\overrightarrow u\:\overrightarrow v\:\overrightarrow w]=?$

1 Answer

$[\overrightarrow u\:\overrightarrow v\:\overrightarrow w]=\overrightarrow u.(\overrightarrow v\times\overrightarrow w)$
$\overrightarrow v\times\overrightarrow w=\left|\begin {array}{ccc}\hat i & \hat j &\hat k\\2 & 1 & -1\\1 & 0 &3\end {array}\right|=3\hat i-7\hat j-\hat k$
$|\overrightarrow v\times\overrightarrow w|=\sqrt {9+49+1}=\sqrt {59}$
$\overrightarrow u.(\overrightarrow v\times\overrightarrow w)=|\overrightarrow u||\overrightarrow v\times\overrightarrow w| cos\theta$ ( where $\theta$ is the angle between $\overrightarrow u\:\:and\:\:\overrightarrow v\times\overrightarrow w)$.
Since $-1\leq cos\theta \leq 1$, the maximum value of $\overrightarrow u.(\overrightarrow v\times\overrightarrow w)$ is
$|\overrightarrow u| |\overrightarrow v\times\overrightarrow w|=\sqrt {59}$ (Since $|\overrightarrow u|=1)$
answered Dec 8, 2013 by rvidyagovindarajan_1
 

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