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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class12  >>  Vector Algebra
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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]$ is ?

$(A)\;-1 \\ (B)\;\sqrt {10}+\sqrt 6\\ (C)\;\sqrt {59}\\(D)\;\sqrt {60} $
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Given: $\overrightarrow v=2\hat i+\hat j-\hat k,\:\:\overrightarrow w=\hat i+3\hat k,\:\:and \:\: |\overrightarrow u|=1$
$[\overrightarrow u\:\overrightarrow v\:\overrightarrow w]=\overrightarrow u.(\overrightarrow v\times\overrightarrow w)$
$=|\overrightarrow u||\overrightarrow v\times\overrightarrow w|\:cos\theta$ (where $\theta$ is angle between $\overrightarrow u$ and $\overrightarrow v\times\overrightarrow w)$
$=|\overrightarrow v\times\overrightarrow w|\:cos\theta$ (since $|\overrightarrow u|=1$.)
This is maximum if $cos\theta=1.$
$\therefore$ The maximum value of $[\overrightarrow u\:\overrightarrow v\:\overrightarrow w]$ is $|\overrightarrow v\times\overrightarrow w|$
$\overrightarrow v\times\overrightarrow w=\left |\begin {array}{ccc} \hat i & \hat j &\hat k \\ 2 & 1 & -1 \\1 & 0 & 3\end {array} \right |=3\hat i-7\hat j-\hat k$
$|\overrightarrow v\times\overrightarrow w|=\sqrt {9+49+1}=\sqrt {59}$
answered Jan 13, 2014 by rvidyagovindarajan_1

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