Question

If $\overrightarrow{\alpha }+\overrightarrow{\beta }+\overrightarrow{\gamma }=a\overrightarrow{\delta }$ and $\overrightarrow{\beta }+\overrightarrow{\gamma }+\overrightarrow{\sigma }=b\overrightarrow{\alpha }$ and $\overrightarrow{\alpha },\overrightarrow{\beta },\overrightarrow{\gamma }$ are not co-planar and $\overrightarrow{\alpha }$ is not parallel to $\overrightarrow{\sigma }$, then $\overrightarrow{\alpha }+\overrightarrow{\beta }+\overrightarrow{\gamma }+\overrightarrow{\sigma }$ equals

• A. $a\overrightarrow{\alpha }$
• B. $b\overrightarrow{\sigma }$
• C. $0$
• D. $(a+b)\overrightarrow{\gamma }$

Answer 1

Answer: (C)

$0$

Given, $\overrightarrow{\alpha }+\overrightarrow{\beta }+\overrightarrow{\gamma }=a\overrightarrow{\sigma }$.....$\left(i\right)$
$\overrightarrow{\beta }+\overrightarrow{\gamma }+\overrightarrow{\sigma }=\overrightarrow{\alpha }$.....$\left(ii\right)$
From Eq. $\left(i\right)$,
$\overrightarrow{\alpha }+\overrightarrow{\beta }+\overrightarrow{\gamma }+\overrightarrow{\sigma }=(a+1)\overrightarrow{\sigma }$...........$\left(iii\right)$
From Eq$\left(ii\right)$,
$\overrightarrow{\alpha }+\overrightarrow{\beta }+\overrightarrow{\gamma }+\overrightarrow{\sigma }=(b+1)\overrightarrow{\alpha }$...........$\left(iv\right)$
From Eqs.$\left(iii\right)$ and $\left(iv\right)$,$(\alpha +1)\overrightarrow{\sigma }=(b+1)\overrightarrow{\alpha }$..............$\left(v\right)$
Since $\overrightarrow{\alpha }$ is not parallel to $\overrightarrow{\sigma }$
$\therefore$ From .$\left(v\right)$,
$a+1=0$ and $b+1=0$
$\therefore$ From Eq.$\left(iii\right)$,
$\overrightarrow{\alpha }+\overrightarrow{\beta }+\overrightarrow{\gamma }+\overrightarrow{\sigma }=0$