Vector Algebra Question 195
Question: Let a, b, c are three non-coplanar vectors such that
$ {{\mathbf{r}}_1}=\mathbf{a}-\mathbf{b}+\mathbf{c},{{\mathbf{r}}_2}=\mathbf{b}+\mathbf{c}-\mathbf{a},{{\mathbf{r}}_3}=\mathbf{c}+\mathbf{a}+\mathbf{b}, $
$ \mathbf{r}=2\mathbf{a}-3\mathbf{b}+4\mathbf{c}. $ If $ \mathbf{r}={\lambda_1}{{\mathbf{r}}_1}+{\lambda_2}{{\mathbf{r}}_2}+{\lambda_3}{{\mathbf{r}}_3}, $ then
Options:
A) $ {\lambda_1}=7 $
B) $ {\lambda_1}+{\lambda_3}=3 $
C) $ {\lambda_1}+{\lambda_2}+{\lambda_3}=4 $
D) $ {\lambda_3}+{\lambda_2}=2 $
Show Answer
Answer:
Correct Answer: B
Solution:
- We have $ \mathbf{r}={\lambda_1}{{\mathbf{r}}_1}+{\lambda_2}{{\mathbf{r}}_2}+{\lambda_3}{{\mathbf{r}}_3} $
$ \Rightarrow 2\mathbf{a}-3\mathbf{b}+4\mathbf{c}=({\lambda_1}-{\lambda_2}+{\lambda_3})\mathbf{a} $
$ +(-{\lambda_1}+{\lambda_2}+{\lambda_3})\mathbf{b}+({\lambda_1}+{\lambda_2}+{\lambda_3})\mathbf{c} $
$ \Rightarrow {\lambda_1}-{\lambda_2}+{\lambda_3}=2,-{\lambda_1}+{\lambda_2}+{\lambda_3}=-3,{\lambda_1}+{\lambda_2}+{\lambda_3}=4 $ $ (\because ,\mathbf{a},,\mathbf{b},,\mathbf{c} $ are non-coplanar)
$ \Rightarrow {\lambda_1}=\frac{7}{2}, $ $ {\lambda_2}=1, $ $ {\lambda_3}=-\frac{1}{2} $
Therefore, $ {\lambda_1}+{\lambda_3}=3 $ and $ {\lambda_1}+{\lambda_2}+{\lambda_3}=4 $ .
BETA
