Question

$\mathbf{p} = 2\mathbf{a} - 3\mathbf{b},\mathbf{q} = \mathbf{a} - 2\mathbf{b} + \mathbf{c},\mathbf{r} = - 3\mathbf{a} + \mathbf{b} + 2\mathbf{c};$ where a, b and c being non-zero, non-coplanar vectors, then the vector $- 2\mathbf{a} + 3\mathbf{b} - \mathbf{c}$ is equal to

• A. $\mathbf{p} - 4\mathbf{q}$
• B. $\frac{- 7\mathbf{q} + \mathbf{r}}{5}$
• C. $2\mathbf{p} - 3\mathbf{q} + \mathbf{r}$
• D. $4\mathbf{p} - 2\mathbf{r}$

Answer 1

Answer: (B)

$\frac{- 7\mathbf{q} + \mathbf{r}}{5}$

Answer 2

Let $- 2\mathbf{a} + 3\mathbf{b} - \mathbf{c} = x\mathbf{p} + y\mathbf{q} + z\mathbf{r}$

$\Rightarrow - 2\mathbf{a} + 3\mathbf{b} - \mathbf{c}$

$= ( 2 x + y - 3 z ) \mathbf { a } + ( - 3 x - 2 y + z ) \mathbf { b } + ( y + 2 z ) \mathbf { c }$

$\therefore 2x + y - 3z = - 2,$ $- 3x - 2y + z = 3$ and $y + 2z = - 1$

Solving these, we get $x = 0,$ $y = - \frac{7}{5},$ $z = \frac{1}{5}$

∴ $- 2\mathbf{a} + 3\mathbf{b} - \mathbf{c} = \frac{( - 7\mathbf{q} + \mathbf{r})}{5}.$

Trick : Check alternates one by one

i.e., (1) $\mathbf{p} - 4\mathbf{q} = - 2\mathbf{a} + 5\mathbf{b} - 4\mathbf{c}$

(2) $\frac{- 7\mathbf{q} + \mathbf{r}}{5} = - 2\mathbf{a} + 3\mathbf{b} - \mathbf{c}$.