Question

If $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are noncoplanar and $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\alpha \overrightarrow{d}$, $\overrightarrow{b}+\overrightarrow{c}+\overrightarrow{d}=\beta \overrightarrow{a}$ then $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}+\overrightarrow{d}=$

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

Answer 1

The correct option is A $0$

$\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\alpha \overrightarrow{d}$(given)
$\overrightarrow{b}+\overrightarrow{c}+\overrightarrow{d}=\beta \overrightarrow{a}$(given)
Eliminating $\overrightarrow{d}$,
$\Rightarrow \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\alpha \overrightarrow{d}$
$\Rightarrow \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\alpha (\beta \overrightarrow{a}-\overrightarrow{b}-\overrightarrow{c})$
$\Rightarrow (1-\alpha \beta )\overrightarrow{a}+(1+\alpha )\overrightarrow{b}+(1+\alpha )\overrightarrow{c}=0$ since $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are non-coplanar
$\Rightarrow 1-\alpha \beta =0,1+\alpha =0$
$\Rightarrow \alpha \beta =1,\alpha =-1$
$\Rightarrow \alpha =-1\therefore \beta =-1$
$\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\alpha \overrightarrow{d}=-\overrightarrow{d}$
$\therefore \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}+\overrightarrow{d}=0$