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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class12  >>  Vector Algebra
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In a rt. angled $\Delta$ $ABC$, the hypotenuse $|\overrightarrow {AB}|=p$,then $\overrightarrow {AB}.\overrightarrow {AC}+\overrightarrow {BC}.\overrightarrow {BA}+\overrightarrow {CA}.\overrightarrow {CB}=?$

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Since $\overrightarrow {AB}$ is hypotenuse, right angle is at $C$
$\Rightarrow\:\overrightarrow {CA}.\overrightarrow {CB}=0$
$\overrightarrow {AB}.\overrightarrow {AC}+\overrightarrow {BC}.\overrightarrow {BA}+\overrightarrow {CA}.\overrightarrow {CB}$
$=\overrightarrow {AB}.\overrightarrow {AC}-\overrightarrow {BC}.\overrightarrow {AB}$
$=\overrightarrow {AB}.(\overrightarrow {AC}-\overrightarrow {BC})=\overrightarrow {AB}.(\overrightarrow {AC}+\overrightarrow {CB})$
$=\overrightarrow {AB}.\overrightarrow {AB}=|\overrightarrow {AB}|^2=p^2$
answered Nov 6, 2013 by rvidyagovindarajan_1

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