Question

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

$\overline{0}$

$\overrightarrow{d}=\frac{\overrightarrow{a}}{x}+\frac{(\overrightarrow{a}+\overrightarrow{c})}{x}$ ...(1)

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

$\Rightarrow \overrightarrow{d}=\overrightarrow{a}y-(\overrightarrow{a}+\overrightarrow{c})$ ...(2)

from (1) and (2)

$\frac{\overrightarrow{a}}{x}+\frac{(\overrightarrow{b}+\overrightarrow{c})}{x}=\overrightarrow{a}y-(\overrightarrow{b}+\overrightarrow{c})$

$\because \overrightarrow{a},\overrightarrow{b}$ and $\overrightarrow{c}$ are non-coplanar

$\frac{1}{x}=y$ and $x=-1\to y=-1$

$\therefore \overrightarrow{d}=\frac{1}{x}(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c})=-(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c})$ ...(3)

$\therefore \frac{1}{xy}(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}+\overrightarrow{d})=1[\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}-(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\left)\right]=\overrightarrow{0}$