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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class12  >>  Vector Algebra
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The angle between the diagonals of a parallelogram $ABCD$, where $\overrightarrow {AB}=3\hat i-2\hat j+2\hat k$ and $\overrightarrow {BC}=-\hat i-2\hat k$ is?

$\begin{array}{1 1} \large\frac{\pi}{6} \\ \end{array} $

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One diagonal is $\overrightarrow {AC}=\overrightarrow {AB}+\overrightarrow{BC}$
i.e.,$\overrightarrow {AC}=2\hat i-2\hat j$
Other diagonal is $\overrightarrow{DB}=\overrightarrow {AB}-\overrightarrow {BC}$
i.e.,$\overrightarrow {DB}=4\hat i-2\hat j+4\hat k$
Angle between the diagonals is given by
$cos^{-1}\bigg[\large\frac{\overrightarrow {AC}.\overrightarrow {DB}}{|\overrightarrow {AC}||\overrightarrow {DB}|}\bigg]$
$=cos^{-1}\bigg[\large\frac{8+4}{\sqrt 8.\sqrt {36}}\bigg]$=$cos^{-1}\large\frac{1}{\sqrt 2}$
answered Nov 1, 2013 by rvidyagovindarajan_1

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