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Questions  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class12  >>  Vector Algebra
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Q)

If $ \overrightarrow x=\large\frac{\overrightarrow b\times\overrightarrow c}{[\overrightarrow a\:\overrightarrow b\:\overrightarrow c]},\:\:\overrightarrow y=\large\frac{\overrightarrow c\times\overrightarrow a}{[\overrightarrow a\:\overrightarrow b\:\overrightarrow c]},\:\:and\:\:\overrightarrow z=\large\frac{\overrightarrow a\times\overrightarrow b}{[\overrightarrow a\:\overrightarrow b\:\overrightarrow c]},$ then $\overrightarrow x.(\overrightarrow a+\overrightarrow b)+\overrightarrow y.(\overrightarrow b+\overrightarrow c)+\overrightarrow z.(\overrightarrow c+\overrightarrow a)=?$

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A)
$\overrightarrow x.(\overrightarrow a+\overrightarrow b)=\large\frac{\overrightarrow b\times\overrightarrow c}{[\overrightarrow a\:\overrightarrow b\:\overrightarrow c]}$$.(\overrightarrow a+\overrightarrow b)$
$=\large\frac{[\overrightarrow b\:\overrightarrow c\:\overrightarrow a]+[\overrightarrow b\:\overrightarrow c\:\overrightarrow b]}{[\overrightarrow a\:\overrightarrow b\:\overrightarrow c]}$
$=\large\frac{[\overrightarrow b\:\overrightarrow c\:\overrightarrow a]}{[\overrightarrow a\:\overrightarrow b\:\overrightarrow c]}=1$
Similarly by the law of symmetricity,
$\overrightarrow y.(\overrightarrow b+\overrightarrow c)=\large\frac{\overrightarrow c\times\overrightarrow a}{[\overrightarrow a\:\overrightarrow b\:\overrightarrow c]}$$.(\overrightarrow b+\overrightarrow c)=1$
and
$\overrightarrow z.(\overrightarrow c+\overrightarrow a)=\large\frac{\overrightarrow c\times\overrightarrow a}{[\overrightarrow a\:\overrightarrow a\:\overrightarrow b]}$$.(\overrightarrow c+\overrightarrow a)=1$
$\therefore\:\overrightarrow x.(\overrightarrow a+\overrightarrow b)+\overrightarrow y.(\overrightarrow b+\overrightarrow c)+\overrightarrow z.(\overrightarrow c+\overrightarrow a)=3$
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