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

In a rt. angled triangle $ABC$ if hypotenuse, $AB=p$, then $\overrightarrow {AB}.\overrightarrow {AC}+\overrightarrow {BC}.\overrightarrow {BA}+\overrightarrow {CA}.\overrightarrow {CB}$= ?

$(a)\:\:2p^2\:\:\:\qquad\:\:(b)\:\:\large\frac{p^2}{2}$$\:\:\:\qquad\:\:(c)\:\:p^2\:\:\:\qquad\:\:(d)\:\:3p^2$

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  • $\overrightarrow a.\overrightarrow a=|\overrightarrow a|^2$
Since $ABC$ is rt. angled $\Delta$ and $AB$ is hypotenuse $<\:C=90^{\circ}$
$\therefore\:\overrightarrow {CA}.\overrightarrow {CB}=0$
$\Rightarrow\:\overrightarrow {AB}.\overrightarrow {AC}+\overrightarrow {BC}.\overrightarrow {BA}+\overrightarrow {CA}.\overrightarrow {CB}=\Rightarrow\:\overrightarrow {AB}.\overrightarrow {AC}+\overrightarrow {BC}.\overrightarrow {BA}+0$
$=\overrightarrow {AB}.\overrightarrow {AC}+\overrightarrow {CB}.\overrightarrow {AB}$
$=\overrightarrow {AB}.(\overrightarrow {AC}+\overrightarrow {CB})$
But from triangular law of addition, $\overrightarrow {AC}+\overrightarrow {CB}=\overrightarrow {AB}$
$=\overrightarrow {AB}.\overrightarrow {AB}=AB^2=p^2$
answered Jan 7, 2014 by rvidyagovindarajan_1
 

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