Question

If $p,q,r$ are three mutually perpendicular vectors of the same magnitude and if a vector $x$ satisfies the equation $p\times \left(\right(x-q)\times p)+q\times \left(\right(x-r)\times q)+r\times \left(\right(x-p)\times r)=0,$ then vector the $x$ is

• A. $\frac{1}{2}(p+q-2r)$
• B. $\frac{1}{2}(p+q+r)$
• C. $\frac{1}{3}(p+q+r)$
• D. $\frac{1}{3}(2p+q-r)$

Answer 1

Answer: (B)

$\frac{1}{2}(p+q+r)$

$p\times \left(\right(x-q)\times p)=(p\cdot p)(x-q)-(p\cdot (x-q\left)\right)p={\left|p\right|}^2(x-q)-(p\cdot x)p,(\because p\cdot q=0)$
Similarly $q\times \left(\right(x-p)\times r)={\left|q\right|}^2(x-r)-(q\cdot x)q$
and $r\times \left(\right(x-p)\times r)={\left|r\right|}^2(x-p)-(r\cdot x)r$
Without loss of generality we can assume that
$p=\left|p\right|i,q=\left|p\right|j,r=\left|p\right|k,(\because \left|p\right|=\left|q\right|=\left|r\right|)$
Therefore L.H.S. of the given expressiion
$=3{\left|p\right|}^{2}x-{\left|p\right|}^2(p+q+r)-{\left|p\right|}^2\left[\right(i\cdot x)i+(j\cdot x)j+(b\cdot x\left)k\right]$
$=3{\left|p\right|}^{2}x-{\left|p\right|}^2(p+q+r)-{\left|p\right|}^{2}x$
$=2{\left|p\right|}^{2}x-{\left|p\right|}^2(p+q+r)$
This expression will be equal to $0$ if $x=\frac{1}{2}(p+q+r)$