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

If $\overrightarrow d=x(\overrightarrow a \times\overrightarrow b)+y(\overrightarrow b\times\overrightarrow c)+z(\overrightarrow c\times\overrightarrow a)$ and $[\overrightarrow a\:\overrightarrow b\:\overrightarrow c]=\large\frac{1}{8}$, then $x+y+z=?$

$\begin{array}{1 1} (A)\frac{3}{8} [\overrightarrow a\:\overrightarrow b\:\overrightarrow c] \\ (B) 8(\overrightarrow a+\overrightarrow b+\overrightarrow c).\overrightarrow d \\ (C) \large\frac{3}{8}(\overrightarrow a+\overrightarrow b+\overrightarrow c).\overrightarrow d \\ (D) \large\frac{1}{8}(\overrightarrow a+\overrightarrow b+\overrightarrow c).\overrightarrow d \end{array} $

1 Answer

$\overrightarrow a.\overrightarrow d=y\:\overrightarrow a.(\overrightarrow b\times\overrightarrow c)=y[\overrightarrow a\:\overrightarrow b\:\overrightarrow c]=\large\frac{1}{8}$$y$
Similarly $\overrightarrow b.\overrightarrow d=\large\frac{1}{8}$$z$ and $\overrightarrow c.\overrightarrow d=\large\frac{1}{8}$$x$
$\therefore\:\large\frac{1}{8}$$(x+y+z)=(\overrightarrow a+\overrightarrow b+\overrightarrow c).\overrightarrow d$
$\Rightarrow\:(x+y+z)=8(\overrightarrow a+\overrightarrow b+\overrightarrow c).\overrightarrow d$
answered Dec 6, 2013 by rvidyagovindarajan_1
 

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