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If $\overrightarrow u,\overrightarrow v,\overrightarrow w$ are three vectors such that $|\overrightarrow u|=1,\:|\overrightarrow v|=2,\:|\overrightarrow w|=3$ and if the projection of $\overrightarrow v$ along $\overrightarrow u$ is equal to that of $\overrightarrow w$ along $\overrightarrow u$ and also if $\overrightarrow u$ and $\overrightarrow v$ are $\perp$ to each other, then $ |\overrightarrow u-\overrightarrow v+\overrightarrow w|= ?$

$\begin{array}{1 1} 14 \\ 7 \\ \sqrt {14} \\ \sqrt {7} \end{array} $

1 Answer

Given: Projection of $\overrightarrow v$ along $\overrightarrow u$ = Projection of $\overrightarrow w$ along $\overrightarrow u$
$\Rightarrow\:\large\frac{\overrightarrow v.\overrightarrow u}{|\overrightarrow u|}=\frac{\overrightarrow w.\overrightarrow u}{|\overrightarrow u|}$
$\Rightarrow\:\overrightarrow v.\overrightarrow u=\overrightarrow w.\overrightarrow u$............(i)
Given that $\overrightarrow v$ and $\overrightarrow w$ are $\perp $ to each other.
$\therefore\:\overrightarrow v.\overrightarrow w=0....(ii)$
$|\overrightarrow u-\overrightarrow v+\overrightarrow w|^2=|\overrightarrow u|^2+|\overrightarrow v|^2+|\overrightarrow w|^2+2(-\overrightarrow u.\overrightarrow v-\overrightarrow v.\overrightarrow w+\overrightarrow w.\overrightarrow u)$
$=1+4+9+2\times( 0)=14$ ( From (i) and (ii))
$\therefore\:|\overrightarrow u-\overrightarrow v+\overrightarrow w|=\sqrt {14}$


answered Nov 18, 2013 by rvidyagovindarajan_1
edited Mar 5, 2014 by meenakshi.p

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