Question

Let $a,b,c$ be three non-coplanar vectors such that $r_1=a-b+c$, $r_2=b+c-a$, $r_3=c+a+b$,$r=2a-3b+4c$. If $r={\lambda }_1{r}_1+{\lambda }_2{r}_2+{\lambda }_3{r}_3$, then

• A. ${\lambda }_1=7$
• B. ${\lambda }_1+{\lambda }_3=6$
• C. ${\lambda }_1+{\lambda }_2+{\lambda }_3=4$
• D. ${\lambda }_2+{\lambda }_3=2$

Answer 1

The correct option is C

${\lambda }_1+{\lambda }_2+{\lambda }_3=4$

Explanation for the correct option:

Simplification of expression:

Given, $r={\lambda }_1{r}_1+{\lambda }_2{r}_2+{\lambda }_3{r}_3$

$\therefore$ $2a-3b+4c={\lambda }_1{r}_1+{\lambda }_2{r}_2+{\lambda }_3{r}_3$

$\Rightarrow$ $2a-3b+4c={\lambda }_1\left(a-b+c\right)+{\lambda }_2\left(b+c-a\right)+{\lambda }_3(c+a+b)$

$\Rightarrow$ $2a-3b+4c=a\left({\lambda }_1-{\lambda }_2+{\lambda }_3\right)+b\left({\lambda }_2-{\lambda }_1+{\lambda }_3\right)+c\left({\lambda }_1+{\lambda }_2+{\lambda }_3\right)$

On comparing the coefficients of $c$ we get,

$\left({\lambda }_1+{\lambda }_2+{\lambda }_3\right)=4$

Hence option (C) is the correct answer.