Question

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

• A. $3c$
• B. $-a$
• C. $0$
• D. $2a$

Answer 1

The correct option is C

$0$

Explanation for the correct option:

Step 1. Find the value of $\frac{1}{xy}(a+b+c+d)$:

Given, $d=\frac{1}{x}(a+b+c)$

$\Rightarrow$ $dx=a+b+c$

$\Rightarrow$$b+c=dx-a$ …..(1)

$a=\frac{1}{y}(b+c+d)$

$\Rightarrow$ $ay=b+c+d$

$\Rightarrow$$b+c=ay-d$ …..(2)

Step 2. From (1) and (2), we get

$dx-a=ay-d$

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

$\therefore d\parallel a$

Step 3. If $a,b,c$ and $d$ are non-coplanar vectors, then

$x=-1$ and $y=-1$

$\begin{array}{rcl}& \Rightarrow & d=\frac{1}{x}(a+b+c)\\ & =& -(a+b+c)\end{array}$

$\begin{array}{rcl}\therefore \frac{1}{xy}(a+b+c+d)& =& 1[a+b+c-(a+b+c\left)\right]\\ & =& \mathbf{0}\end{array}$

Hence, option ‘C’ is Correct.