Vector Algebra Question 209
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
Options:
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} $
Show Answer
Answer:
Correct Answer: B
Solution:
- 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} $ $ =(2x+y-3z)\mathbf{a}+(-3x-2y+z)\mathbf{b}+(y+2z)\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., (a) $ \mathbf{p}-4\mathbf{q}=-2\mathbf{a}+5\mathbf{b}-4\mathbf{c} $ (b) $ \frac{-7\mathbf{q}+\mathbf{r}}{5}=-2\mathbf{a}+3\mathbf{b}-\mathbf{c} $ .
BETA
