Vector Algebra Question 167
Question: If $ \mathbf{a}\times \mathbf{r}=\mathbf{b}+\lambda \mathbf{a} $ and $ \mathbf{a}.\mathbf{r}=3, $ where $ \mathbf{a}=2\mathbf{i}+\mathbf{j}-\mathbf{k} $ and $ \mathbf{b}=-\mathbf{i}-2\mathbf{j}+\mathbf{k}, $ then r and l are equal to
Options:
A) $ \mathbf{r}=\frac{7}{6}\mathbf{i}+\frac{2}{3}\mathbf{j},\lambda =\frac{6}{5} $
B) $ \mathbf{r}=\frac{7}{6}\mathbf{i}+\frac{2}{3}\mathbf{j},\lambda =\frac{5}{6} $
C) $ \mathbf{r}=\frac{6}{7}\mathbf{i}+\frac{2}{3}\mathbf{j},\lambda =\frac{6}{5} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
- Given, $ \mathbf{a}\times \mathbf{r}=\mathbf{b}+\lambda ,\mathbf{a}\Rightarrow (\mathbf{a}\times \mathbf{r}),.,\mathbf{a}=\mathbf{b},.,\mathbf{a}+\lambda ,\mathbf{a},.,\mathbf{a} $
$ \Rightarrow 0=\mathbf{b},.,\mathbf{a}+\lambda ,|\mathbf{a}{{|}^{2}}\Rightarrow \lambda =-\frac{\mathbf{b},.\mathbf{a}}{|\mathbf{a}{{|}^{2}}}=\frac{5}{6} $
Also, $ (\mathbf{a}\times \mathbf{r})\times \mathbf{a}=\mathbf{b}\times \mathbf{a}+\lambda ,\mathbf{a}\times \mathbf{a}\Rightarrow \mathbf{r}=\frac{7}{6}\mathbf{i}+\frac{2}{3}\mathbf{j} $ .
BETA
