Question

If $\overrightarrow a ,\overrightarrow b $ and $\overrightarrow c $ are non – coplanar vectors and if $\overrightarrow d $ is such that $\overrightarrow d = \dfrac{1}{x}\left( {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right)$ and $\overrightarrow a = \dfrac{1}{y}\left( {\overrightarrow b + \overrightarrow c + \overrightarrow d } \right)$ , where $x$ and $y$ are non – zero real numbers, then $\dfrac{1}{{xy}}\left( {\overrightarrow a + \overrightarrow b + \overrightarrow c + \overrightarrow d } \right)$ equals ?
1. $3\overrightarrow c $
2. $ - \overrightarrow a $
3. $0$
4. $2\overrightarrow a $

Answer 1

Hint: Solve the given values of vector $\overrightarrow d $ and $\overrightarrow a $ then compare their values. Apply non – coplanar vectors condition to find the value of $x$ and $y$. In last, put all the required values in $\dfrac{1}{{xy}}\left( {\overrightarrow a + \overrightarrow b + \overrightarrow c + \overrightarrow d } \right)$ and get the value.

Formula Used:
$\overrightarrow d = \dfrac{1}{x}\left( {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right)$,$\overrightarrow a = \dfrac{1}{y}\left( {\overrightarrow b + \overrightarrow c + \overrightarrow d } \right)$(Given)

Complete step by step Solution:
Given that,
$\overrightarrow d = \dfrac{1}{x}\left( {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right)$
$ \Rightarrow \overrightarrow d x = \overrightarrow a + \overrightarrow b + \overrightarrow c $
$\overrightarrow b + \overrightarrow c = \overrightarrow d x - \overrightarrow a - - - - - (1)$
And $\overrightarrow a = \dfrac{1}{y}\left( {\overrightarrow b + \overrightarrow c + \overrightarrow d } \right)$
$ \Rightarrow \overrightarrow a y = \overrightarrow b + \overrightarrow c + \overrightarrow d $
$\overrightarrow b + \overrightarrow c = \overrightarrow a y - \overrightarrow d - - - - - (2)$
From equation (1) and (2),
$\overrightarrow d x - \overrightarrow a = \overrightarrow a y - \overrightarrow d $
$\overrightarrow d x + \overrightarrow d = \overrightarrow a y + \overrightarrow a $
$\overrightarrow d \left( {x + 1} \right) = \overrightarrow a (y + 1)$
$ \Rightarrow \overrightarrow d \parallel \overrightarrow a $
Since $\overrightarrow a ,\overrightarrow b ,\overrightarrow c $ and $\overrightarrow d $ are non – coplanar vectors.
Therefore, $x = - 1,y = - 1$
Put the values in $\overrightarrow d = \dfrac{1}{x}\left( {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right)$
So, $\overrightarrow d = \dfrac{1}{{ - 1}}\left( {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right)$
$\overrightarrow d = - \overrightarrow a - \overrightarrow b - \overrightarrow c - - - - - (3)$
Now, $\dfrac{1}{{xy}}\left( {\overrightarrow a + \overrightarrow b + \overrightarrow c + \overrightarrow d } \right)$
$ = \dfrac{1}{{\left( { - 1} \right)\left( { - 1} \right)}}\left( {\overrightarrow a + \overrightarrow b + \overrightarrow c + \left( { - \overrightarrow a - \overrightarrow b - \overrightarrow c } \right)} \right)Equation{\text{ }}\left( 3 \right)$
$ = 0$

Hence, the correct option is 3.

Note:The key concept involved in solving this problem is a good knowledge of non-coplanar vectors. Students must remember that Vectors are said to be non-coplanar if and only if their support lines are not parallel to the same plane.