Question

If $\overline{a},\overline{b},\overline{c}$ are non-coplanar vectors and if $\overline{d}$ is such that $\overline{d}=\frac{1}{x}(\overline{a}+\overline{b}+\overline{c})$ and $\overline{d}=\frac{1}{y}(\overline{b}+\overline{c}+\overline{d})$ where x and y are non-zero real numbers, then $\frac{1}{xy}(\overline{a}+\overline{b}+\overline{c}+\overline{d})=$

• A. $3\overline{c}$
• B. $-\overline{a}$
• C. $\overline{0}$
• D. $2\overline{a}$

Answer 1

Answer: (C)

$\overline{0}$

Given

$\overrightarrow{d}=\frac{\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}}{x}$

$dx=a+b+c$

$\Rightarrow b+c=dx-a$ ___ (i)

$\overrightarrow{a}=\frac{\overrightarrow{b}+\overrightarrow{c}+\overrightarrow{d}}{y}$

$\Rightarrow b+c=ay-d$ ___ (ii)

$\therefore dx-a=ay-d$

$\Rightarrow (x+1)d=(y+1)a$

$=\overrightarrow{d}\parallel \overrightarrow{a}$

Also If is given that $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c},\overrightarrow{d}$ are non-coplanar vector

$\therefore$ It is possible when

$x=-1,y=-1$

$\therefore \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}+\overrightarrow{d}=\overrightarrow{0}$

$\frac{\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}+\overrightarrow{d}}{xy}=\frac{\overrightarrow{0}}{(-1)(-1)}=\overrightarrow{0}$