Let $\vec{a}, \vec{b}$ and $\vec{c}$ be any three non coplanar vectors. If $m$, $n$ are scalars such that…
Let $\vec{a}, \vec{b}$ and $\vec{c}$ be any three non coplanar vectors. If $m$, $n$ are scalars such that $\vec{a}+\vec{b}=m \vec{d}-\vec{c}$ and $\vec{b}+\vec{c}=n \vec{a}-\vec{d}$, then $3 \vec{a}+2 \vec{b}+2 \vec{c}+\vec{d}=$
$\vec{a}-\vec{d}$
$\vec{a}+\vec{d}$
$\overrightarrow{0}$
$\vec{b}+\vec{c}+2 \vec{d}$
Solution
$\because \vec{a}+\vec{b}=m \vec{d}-\vec{c} \Rightarrow \vec{a}+\vec{b}+\vec{c}=m \vec{d}...(i)$
$\& \vec{b}+\vec{c}=n \vec{a}-\vec{d} \Rightarrow \vec{d}=n \vec{a}+(-\vec{b}+\vec{c})...(ii)$
From equations (i) \& (ii), we get
$\begin{aligned}
& (\vec{a}+\vec{b}+\vec{c})=m(n \vec{a}-\vec{b}-\vec{c}) \\
& \Rightarrow(1-m n) \vec{a}+(1+m) \vec{b}+(1+m) \vec{c}=0
\end{aligned}$
$\vec{a}, \vec{b}$ and $\vec{c}$ are non-coplaner vectors
$\therefore 1=m n \text { and } 1+m=0 \Rightarrow m=-1$
$\therefore n=-1$
Putting the value of $m$ in equation (1)
$\vec{a}+\vec{b}+\vec{c}=-\vec{d} \Rightarrow \vec{a}+\vec{b}+\vec{c}+\vec{d}=0$
Now, $3 \vec{a}+2 \vec{b}+2 \vec{c}+\vec{d}=(\vec{a}+\vec{b}+\vec{c}+\vec{d})+(\vec{a}+\vec{b}+\vec{c})+\vec{a}$
$=0+(-\vec{d})+\vec{a}=\vec{a}-\vec{d}$