logo

Ask Questions, Get Answers

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

If $\overrightarrow a,\overrightarrow b$ are mutually $\perp$ unit vectors and the vectors $x\overrightarrow a+x\overrightarrow b+z(\overrightarrow a\times\overrightarrow b ),\:\:\overrightarrow a+(\overrightarrow a\times\overrightarrow b),\:\:and\:\:z\overrightarrow a+z\overrightarrow b+y(\overrightarrow a\times\overrightarrow b)$ lie in a plane then $Z$ is ?

$\begin {array} {1 1} (A)\;A.M \: of \: x \: and \: y & \quad (B)\;G.M \: of \: x \: and \: y \\ (C)\;H.M \: of \: x \: and \: y & \quad (D)\;Equal \: to \: zero \end {array}$

Download clay6 mobile app

1 Answer

Given that $\overrightarrow a\:and\:\overrightarrow b$ are mutually $\perp$ unit vectors.
$\Rightarrow\:\overrightarrow a\times\overrightarrow b\neq 0$
Also given that $x\overrightarrow a+x\overrightarrow b+z(\overrightarrow a\times\overrightarrow b),\:\overrightarrow a+(\overrightarrow a\times\overrightarrow b),\:and\:\:z\overrightarrow a+z\overrightarrow b+y(\overrightarrow a\times\overrightarrow b)$ are coplanar.
$\Rightarrow\:[x\overrightarrow a+x\overrightarrow b+z(\overrightarrow a\times\overrightarrow b)\:\overrightarrow a+(\overrightarrow a\times\overrightarrow b)\:\:\:z\overrightarrow a+z\overrightarrow b+y(\overrightarrow a\times\overrightarrow b)]=0$
$\Rightarrow\:\left |\begin {array}{ccc} x & x & z\\ 1 & 0 & 1\\ z & z & y\end {array}\right|.[\overrightarrow a\:\overrightarrow b\:\overrightarrow a\times\overrightarrow b]=0$
$\Rightarrow\:(z^2-xy).\big[(\overrightarrow a\times \overrightarrow b).(\overrightarrow a \times\overrightarrow b)\big]=0$
$\Rightarrow\:(z^2-xy)|\overrightarrow a\times\overrightarrow b|^2=0$
Since $ |\overrightarrow a\times\overrightarrow b|\neq 0$, $z^2-xy=0$
$\Rightarrow\:z$ is $G.M.\:of\:x\:and\:y$
answered Jan 6, 2014 by rvidyagovindarajan_1
edited Mar 25, 2014 by sreemathi.v
 

Related questions

...
X